1. List the department IDs for departments that do not contain the job ID ST_CLERK, using SET operators. SELECT department_id FROM departments MINUS SELECT department_id FROM employees WHERE job_id = ’ST_CLERK’; 2. Display the country ID and the name of the countries that have no departments located in them, using SET operators. SELECT country_id,country_name FROM countries MINUS SELECT l.country_id,c.country_name FROM locations l, countries c WHERE l.country_id = c.country_id; 3. Produce a list of jobs for departments 10, 50, and 20, in that order. Display job ID and department ID, using SET operators. COLUMN dummy NOPRINT SELECT job_id, department_id, ’x’ dummy FROM employees WHERE department_id = 10 UNION SELECT job_id, department_id, ’y’ FROM employees WHERE department_id = 50 UNION SELECT job_id, department_id, ’z’ FROM employees WHERE department_id = 20 ORDER BY 3; COLUMN dummy PRINT 4. List the employee IDs and job IDs of those employees who currently have the job title that they held before beginning their tenure with the company. SELECT employee_id,job_id FROM employees INTERSECT SELECT employee_id,job_id FROM job_history; 5. Write a compound query that lists the following: • Last names and department ID of all the employees from the EMPLOYEES table, regardless of whether or not they belong to any department • Department ID and department name of all the departments from the DEPARTMENTS table, regardless of whether or not they have employees working in them SELECT last_name,department_id,TO_CHAR(null) FROM employees UNION SELECT TO_CHAR(null),department_id,department_name FROM departments;