diff --git a/sql/task3.sql b/sql/task3.sql
index f078a9439..3ae738dce 100644
--- a/sql/task3.sql
+++ b/sql/task3.sql
@@ -2,18 +2,69 @@
 -- Write an SQL query to retrieve the top 3 categories with the highest total sales amount.
 -- The result should include the category ID, category name, and the total sales amount.
 -- Hint: You may need to use subqueries, joins, and aggregate functions to solve this problem.
+-- Selecting the category ID, category name, and the total sales amount for the top 3 categories
+SELECT c.category_id, c.category_name, SUM(oi.total_price) AS total_sales_amount
+FROM Categories c
+-- Joining the 'Products' table to get product details
+JOIN Products p ON c.category_id = p.category_id
+-- Joining the 'Order_Items' table to get order item details
+JOIN Order_Items oi ON p.product_id = oi.product_id
+-- Grouping the results by category_id to calculate the total sales amount per category
+GROUP BY c.category_id, c.category_name
+-- Ordering the results by total_sales_amount in descending order to get the highest sales at the top
+ORDER BY total_sales_amount DESC
+-- Limiting the results to the top 3 categories
+LIMIT 3;
+
 
 -- Problem 10: Retrieve the users who have placed orders for all products in the Toys & Games
 -- Write an SQL query to retrieve the users who have placed orders for all products in the Toys & Games
 -- The result should include the user ID and username.
 -- Hint: You may need to use subqueries, joins, and aggregate functions to solve this problem.
 
+-- Selecting the user ID and username for users who have placed orders for all products in 'Toys & Games'
+SELECT u.user_id, u.username
+FROM Users u
+-- Joining the Orders table to get order details
+JOIN Orders o ON u.user_id = o.user_id
+-- Joining the 'Order_Items' table to get order item details
+JOIN Order_Items oi ON o.order_id = oi.order_id
+-- Joining the Products table to get product details for Toys & Games
+JOIN Products p ON oi.product_id = p.product_id AND p.category_id = 'Toys & Games'
+-- Grouping the results by user_id to count unique products ordered by each user
+GROUP BY u.user_id, u.username
+-- Having count of distinct products equal to the total number of products in 'Toys & Games'
+HAVING COUNT(DISTINCT p.product_id) = (SELECT COUNT(*) FROM Products WHERE category_id = 'Toys & Games');
+
+
 -- Problem 11: Retrieve the products that have the highest price within each category
 -- Write an SQL query to retrieve the products that have the highest price within each category.
 -- The result should include the product ID, product name, category ID, and price.
 -- Hint: You may need to use subqueries, joins, and window functions to solve this problem.
 
+-- Selecting the product ID, product name, category ID, and price for products with the highest price in each category
+SELECT p.product_id, p.product_name, p.category_id, p.price
+FROM Products p
+-- Using RANK to assign ranks to products based on price within each category
+-- PARTITION BY is used to restart the ranking for each category
+-- The outer query then filters for products with rank 1 (highest price) within each category
+WHERE RANK() OVER (PARTITION BY p.category_id ORDER BY p.price DESC) = 1;
+
+
 -- Problem 12: Retrieve the users who have placed orders on consecutive days for at least 3 days
 -- Write an SQL query to retrieve the users who have placed orders on consecutive days for at least 3 days.
 -- The result should include the user ID and username.
 -- Hint: You may need to use subqueries, joins, and window functions to solve this problem.
+
+-- Selecting the user ID and username for users who have placed orders on consecutive days for at least 3 days
+SELECT DISTINCT u.user_id, u.username
+FROM Users u
+-- Join the 'Orders' table to get order details
+JOIN Orders o ON u.user_id = o.user_id
+-- Using LAG to compare each order date to the previous one per user
+-- PARTITION BY is used to restart the comparison for each user
+-- The outer query then filters for consecutive orders (difference in days = 1) for at least 3 days
+WHERE DATEDIFF(o.order_date, LAG(o.order_date) OVER (PARTITION BY u.user_id ORDER BY o.order_date)) = 1
+  AND DATEDIFF(o.order_date, LAG(o.order_date, 2) OVER (PARTITION BY u.user_id ORDER BY o.order_date)) = 2;
+
+