diff --git a/sql/task2.sql b/sql/task2.sql
index ad2596731..fc12e5bb2 100644
--- a/sql/task2.sql
+++ b/sql/task2.sql
@@ -3,17 +3,70 @@
 -- The result should include the product ID, product name, and the average rating.
 -- Hint: You may need to use subqueries or common table expressions (CTEs) to solve this problem.
 
+-- Select the product ID, product name, and the average rating of reviews for each product
+SELECT p.product_id, p.product_name, AVG(r.rating) AS average_rating
+FROM Products p
+-- Joining the products and the reviews table on product_id
+JOIN Reviews r ON p.product_id = r.product_id
+-- Grouping the results by product_id to get the average rating per product
+GROUP BY p.product_id, p.product_name
+-- Ordering the results by average_rating in descending order to get the highest ratings at the top
+ORDER BY average_rating DESC;
+
+
+
+
 -- Problem 6: Retrieve the users who have made at least one order in each category
 -- Write an SQL query to retrieve the users who have made at least one order in each category.
 -- The result should include the user ID and username.
 -- Hint: You may need to use subqueries or joins to solve this problem.
 
+-- Selecting the user ID and username from the Users table
+SELECT DISTINCT u.user_id, u.username
+FROM Users u
+-- Joining the orders table to get the orders for each user
+JOIN Orders o ON u.user_id = o.user_id
+-- Joining the order_items table to get the items within those orders
+JOIN Order_Items oi ON o.order_id = oi.order_id
+-- Joining the products' table to get the category of those items
+JOIN Products p ON oi.product_id = p.product_id
+-- Grouping the results by user_id and category_id to count orders per category per user
+GROUP BY u.user_id, u.username, p.category_id
+-- Having count of orders >= 1 for each category ensures at least one order per category
+HAVING COUNT(DISTINCT p.category_id) >= (SELECT COUNT(*) FROM Categories);
+
+
 -- Problem 7: Retrieve the products that have not received any reviews
 -- Write an SQL query to retrieve the products that have not received any reviews.
 -- The result should include the product ID and product name.
 -- Hint: You may need to use subqueries or left joins to solve this problem.
 
+-- Select product ID and product name from the Products table
+SELECT p.product_id, p.product_name
+FROM Products p
+--  Joining the reviews table to find products without reviews
+LEFT JOIN Reviews r ON p.product_id = r.product_id
+-- Where there are no reviews, the review_id will be NULL
+WHERE r.review_id IS NULL;
+
+
+
 -- Problem 8: Retrieve the users who have made consecutive orders on consecutive days
 -- Write an SQL query to retrieve the users who have made consecutive orders on consecutive days.
 -- The result should include the user ID and username.
--- Hint: You may need to use subqueries or window functions to solve this problem.
\ No newline at end of file
+-- Hint: You may need to use subqueries or window functions to solve this problem.
+
+SELECT user_id, username
+-- Using LAG to compare each order date to the previous one per user
+FROM (
+    SELECT u.user_id, u.username, o.order_date,
+           LAG(o.order_date) OVER (PARTITION BY u.user_id ORDER BY o.order_date) AS prev_order_date
+    FROM Users u
+    JOIN Orders o ON u.user_id = o.user_id
+) AS subquery
+  -- filtering for those where the current order date is one day after the previous order date (consecutive orders)
+
+WHERE order_date = prev_order_date + INTERVAL '1 day'
+  -- grouping the results by user_id to ensure there are only users with more than one consecutive order.
+GROUP BY user_id, username
+HAVING COUNT(*) > 1;