diff --git a/sql/task_1.sql b/sql/task_1.sql
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index 000000000..cca16436e
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+++ b/sql/task_1.sql
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+-- Problem 1: Retrieve all products in the Sports category
+-- Assigning the product data table an alias and selecting all the columns that need to be retrieved from the product data table 
+-- Assigning the category data table an alias
+-- Joining the two tables together based on the column name that is identical in both product data table and category data table
+-- Specifying the category name to retrieve all the products from that category
+
+SELECT pd.product_id, pd.product_name, pd.description, pd.price, cd.category_name FROM product_data pd
+JOIN category_data cd ON pd.category_id = cd.category_id
+WHERE cd.category_name = 'Sports & Outdoors';
+
+-- Problem 2: Retrieve the total number of orders for each user
+-- Assigning the user data table an alias and selecting all the columns that need to be retrieved, from the user data table 
+-- Assigning the order data table an alias and renaming column order_id as total_orders
+-- Used COUNT to count the total number of orders for each user from the order data table
+-- Joining the two tables together based on the column name that is identical in both user data table and order data table
+
+SELECT ud.user_id, ud.username, COUNT(od.order_id) AS total_orders FROM user_data ud
+JOIN order_data od ON ud.user_id = od.user_id
+GROUP BY ud.user_id, ud.username, od.order_id
+
+-- Problem 3: Retrieve the average rating for each product
+-- Assigning the product data table an alias and selecting all the columns that need to be retrieved from the product data table
+-- Assigning the review data table an alias and renaming column rating as average_rating
+-- Used AVG to average the ratings for each product from the review data table
+-- Joining the two tables together based on the column name that is identical in both product data table and review data table
+  
+SELECT pd.product_id, pd.product_name, AVG(rd.rating) AS average_rating FROM product_data pd
+JOIN review_data rd ON pd.product_id = rd.product_id
+GROUP BY pd.product_id, pd.product_name, rd.rating
+
+-- Problem 4: Retrieve the top 5 users with the highest total amount spent on orders
+-- Assigning the user data table an alias and selecting all the columns that need to be retrieved from the user data table 
+-- Assigning the order data table an alias
+-- Joining the two tables together based on the column name that is identical in both product data table and category data table
+-- Ordering the total_amount column in the table in descending order and displaying the top 5 highest total amounts
+SELECT ud.user_id, ud.username, od.total_amount FROM user_data ud
+JOIN order_data od ON ud.user_id = od.user_id
+GROUP BY ud.user_id, ud.username, od.total_amount
+ORDER BY od.total_amount DESC
+LIMIT 5;
diff --git a/sql/task_2.sql b/sql/task_2.sql
new file mode 100644
index 000000000..f28e0e65d
--- /dev/null
+++ b/sql/task_2.sql
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+-- Problem 5: Retrieve the products with the highest average rating
+-- Assigning the product data table an alias and selecting all the columns that need to be retrieved from the product data table 
+-- Assigning the review data table an alias and renaming column rating as average rating
+-- Used AVG to average the ratings for each product from the review data table
+-- Joining the two tables together based on the column name that is identical in both product data table and review data table
+-- Ordering the average_rating column in the table in descending order and displaying the top 5 highest average ratings
+
+SELECT pd.product_id, pd.product_name, AVG(rd.rating) AS average_rating FROM product_data pd
+JOIN review_data rd ON pd.product_id = rd.product_id
+GROUP BY pd.product_id, pd.product_name
+ORDER BY average_rating DESC
+LIMIT 5;
+
+-- Problem 6: Retrieve the users who have made at least one order in each category
+-- Assigning the user data table an alias and selecting all the columns that need to be retrieved from the user data table 
+-- Assigning the order data table an alias
+-- Joining two tables together based on the column name that is identical in both user data table and order data table
+-- Assigning the order items data table an alias
+-- Joining two tables together based on the column name that is identical in both order data table and order items data table
+
+SELECT ud.user_id, ud.username FROM user_data ud
+JOIN order_data od ON ud.user_id = od.user_id
+JOIN order_items_data oid ON od.order_id = oid.order_id
+GROUP BY ud.user_id, ud.username
+
+-- Problem 7: Retrieve the products that have not received any reviews
+-- Assigning the product data table an alias and selecting all the columns that need to be retrieved from the product data table 
+-- Assigning the review data table an alias
+-- Joining the two tables together based on the column name that is identical in both product data table and review data table
+-- Checking the condition where the review text column and rating column have no value
+
+SELECT pd.product_id, pd.product_name FROM product_data pd
+JOIN review_data rd ON pd.product_id = rd.product_id
+WHERE rd.review_text IS NULL AND rd.rating IS NULL;
+
+-- Problem 8: Retrieve the users who have made consecutive orders on consecutive days
+-- Assigning the user data table an alias and selecting all the columns that need to be retrieved from the user data table 
+-- Assigning the order data table an alias
+-- Joining two tables together based on the column name that is identical in both user data table and order data table
+-- Again joining two tables together, with a different alias, and same user id, to show the comparison of the user's orders between two dates
+-- HAVING COUNT is used to ensure the user has consecutive orders on consecutive days
+
+SELECT ud.user_id, ud.username FROM user_data ud
+JOIN order_data od ON ud.user_id = od.user_id
+JOIN order_data odtwo ON ud.user_id = odtwo.user_id AND odtwo.order_date = DATE_ADD(od.order_date, INTERVAL 1 DAY)
+GROUP BY ud.user_id, ud.username
+HAVING COUNT(DISTINCT od.order_id) > 1;
diff --git a/sql/task_3.sql b/sql/task_3.sql
new file mode 100644
index 000000000..56f21d9ab
--- /dev/null
+++ b/sql/task_3.sql
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+-- Problem 9: Retrieve the top 3 categories with the highest total sales amount
+-- Assigning the category data table an alias and selecting all the columns that need to be retrieved from the category data table
+-- Multiplying the quantity with price from the order items data table to get the total sales amount
+-- Assigning the product data table an alias
+-- Joining the two tables together based on the column name that is identical in both product data table and category data table
+-- Assigning the order items data table an alias
+-- Joining the two tables together based on the column name that is identical in both order items data table and product data table
+-- Ordering the total sales amount column in the table in descending order and displaying the top 3 highest total sales amount
+
+SELECT cd.category_id, cd.category_name, SUM(oid.quantity*oid.unit_price) AS total_sales_amount FROM category_data cd
+JOIN product_data pd ON cd.category_id = pd.category_id
+JOIN order_items_data oid ON pd.product_id = oid.product_id
+GROUP BY cd.category_id, cd.category_name
+ORDER BY total_sales_amount DESC
+LIMIT 3;
+
+-- Problem 10: Retrieve the users who have placed orders for all products in the Toys & Games
+-- Assigning the user data table an alias and selecting all the columns that need to be retrieved from the user data table
+-- Assigning the order data table an alias
+-- Joining the two tables together based on the column name that is identical in both user data table and order data table
+-- Assigning the order items data table an alias
+-- Joining the two tables together based on the column name that is identical in both order data table and order items data table
+-- Assigning the product data table an alias
+-- Joining the two tables together based on the column name that is identical in both order items data table and product data table
+-- Assigning the category data table an alias
+-- Joining the two tables together based on the column name that is identical in both product data table and category data table
+-- Specifying the category name to retrieve all the products from that category
+
+SELECT ud.user_id, ud.username FROM user_data ud
+JOIN order_data od ON ud.user_id = od.user_id
+JOIN order_items_data oid ON od.order_id = oid.order_id
+JOIN product_data pd ON oid.product_id = pd.product_id
+JOIN category_data cd ON pd.category_id = cd.category_id
+WHERE cd.category_name = 'Toys & Games'
+GROUP BY ud.user_id, ud.username
+
+-- Problem 11: Retrieve the products that have the highest price within each category
+-- Make a CTE for the Highest priced products in each category
+-- Window function is used to create a row number that will be displayed in descending order
+-- Selecting all the columns that need to be displayed from the CTE
+-- Row number stores the highest priced product from each category
+  
+WITH HighestPricedProducts AS (
+  SELECT product_id, product_name, category_id, price, ROW_NUMBER() OVER (PARTITION BY category_id ORDER BY price DESC) AS row_numbers 
+  FROM product_data
+)
+SELECT product_id, product_name, category_id, price 
+FROM HighestPricedProducts
+WHERE row_numbers = 1;
+  
+
+-- Problem 12: Retrieve the users who have placed orders on consecutive days for at least 3 days
+-- Assigning the user data table an alias and selecting all the columns that need to be retrieved from the user data table
+-- Assigning the order data table an alias
+-- Joining the two tables together based on the column name that is identical in both user data table and order data table
+-- Joining two tables together for the second time, with a different alias, and same user id, to show the comparison of the user's orders between the first and second day
+-- Joining two tables together for the third time, with a different alias, and same user id, to show the comparison of the user's orders between the second and third day
+-- HAVING COUNT is used to ensure the user has consecutive orders on consecutive days
+
+SELECT ud.user_id, ud.username FROM user_data ud
+JOIN order_data od ON ud.user_id = od.user_id
+JOIN order_data odtwo ON ud.user_id = odtwo.user_id AND odtwo.order_date = DATE_ADD(od.order_date, INTERVAL 1 DAY)
+JOIN order_data odthree ON ud.user_id = odthree.user_id AND odthree.order_date = DATE_ADD(odtwo.order_date, INTERVAL 2 DAY)
+GROUP BY ud.user_id, ud.username
+HAVING COUNT(DISTINCT od.order_id) > 2;