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+USE sakila;
+SHOW TABLES;
+SELECT COUNT(*) FROM film;
+
+-- Challenge 1.1 : Display all available tables in the Sakila database.
+SHOW TABLES;
+
+
+-- Challenge 1.2 : Retrieve all the data from the tables `actor`, `film` and `customer`.SHOW * FROM actor;
+SELECT * FROM actor;
+SELECT * FROM film;
+SELECT * FROM customer;
+
+
+-- Challenge 1.3 : Retrieve the following columns from their respective tables:
+DESCRIBE film;
+DESCRIBE language;
+DESCRIBE staff;
+-- Challenge 1.3.1 : Titles of all films from the `film` table
+SELECT DISTINCT title FROM film;
+-- Challenge 1.3.2 : List of languages used in films, with the column aliased as `language` from the `language` table
+SELECT name AS language FROM language;
+-- Challenge 1.3.3 : List of first names of all employees from the `staff` table
+SELECT first_name FROM staff;
+
+
+-- Challenge 1.4 : Retrieve unique release years.
+SELECT DISTINCT release_year FROM film;
+
+
+-- Challenge 1.5 : Counting records for database insights:
+-- Challenge 1.5.1 : Determine the number of stores that the company has.
+SELECT COUNT(*) AS `Amount of stores` FROM store;
+-- Challenge 1.5.2 : Determine the number of employees that the company has.
+SELECT COUNT(*) AS `Amount of employees` FROM staff;
+-- Challenge 1.5.3 : Determine how many films are available for rent and how many have been rented.
+-- Part one, available for rent:
+SELECT COUNT(*) AS `Available for rent` FROM inventory;
+-- (I think this code is not enough, but we havent learned how to do it properly yet. 
+-- This counts all that are in the inventory. Also the once that are out rented at the moment.
+-- I talked with Chat GPT about it and it gave me this code:
+SELECT COUNT(*) AS currently_available
+FROM inventory i
+LEFT JOIN rental r
+  ON r.inventory_id = i.inventory_id
+ AND r.return_date IS NULL
+WHERE r.rental_id IS NULL;
+-- But like i said, its not something I could have come up with by myself)
+-- Part two, have been rented:
+SELECT COUNT(*) AS `Have been rented` FROM rental;
+-- (Little extra)
+SELECT COUNT(*) AS `Currently rented` FROM rental WHERE return_date IS NULL;
+-- Challenge 1.5.4 : Determine the number of distinct last names of the actors in the database.
+SELECT COUNT(DISTINCT last_name) AS `Actors last names` FROM actor;
+
+
+-- Challenge 1.6 : Retrieve the 10 longest films.
+SELECT length, title FROM film ORDER BY length DESC LIMIT 10;
+-- (not sure if the title was also asked for, but made more sence to me to also show it)
+
+
+-- Challenge 1.7 : Use filtering techniques in order to:
+-- Challenge 1.7.1 : Retrieve all actors with the first name "SCARLETT".
+SELECT first_name, last_name FROM actor WHERE first_name = 'SCARLETT';
+-- Challenge 1.7.2 : Retrieve all movies that have ARMAGEDDON in their title and have a duration longer than 100 minutes.
+SELECT title, length
+FROM film
+WHERE title LIKE '%ARMAGEDDON%'
+	AND length > 100;
+-- Challenge 1.7.3 : Determine the number of films that include Behind the Scenes content
+SELECT COUNT(special_features)
+FROM film
+WHERE special_features LIKE '%Behind the Scenes%';
+
+
+
+
+
+
+
+
+
+