Query-Sql Developer - oracle-sqldeveloper

Question

I am creating some queries for my project, but I face some difficulties with the follow ones:
A SELECT statement containing a subquery to retrieve a list of Locations (location id and street_address) that have employees with higher salary than the average of their department. The list must contain the number of those employees and their total salary per location. Name these aggregates respectively "emp" and "totalsalary". The locations in the list must be ordered by location_id.
Select LOCATION_ID, STREET_ADDRESS
from HR.LOCATIONS IN
(Select Employee_id
from HR.Employees
Where Salary > round(avg(SALARY)))
order by location_id;
error: SQL command not properly ended
and the second query is the following
The JOB_HISTORY table can contain more than one entries for an employee who was hired more than once. Create a query to retrieve a list of Employees that were hired more than once. Include the columns EMPLOYEE_ID, LAST_NAME, FIRST_NAME and the aggregate "Times Hired".
SELECT FIRST_NAME,LAST_NAME,EMPLOYEE_ID,
count (*)as TIMES_HIRED
from HR.JOB_HISTORY, HR.EMPLOYEES
where EMPLOYEE_ID= LAST_NAME
having COUNT(*) >1;
error: not a single-group

Answer 1

Try these hope they help. I am making an assumption that employee table has Location_Id column. I am adding Employee_id to Group by to make sure you get correct TotalSalary:
Select LOCATION_ID, STREET_ADDRESS, Count(Employee_id) AS emp, SUM(salary) AS totalsalary
from HR.LOCATIONS INNER JOIN
(Select Employee_id, salary
from HR.Employees
Having Salary > round(avg(SALARY), 0)) AS Emp ON HR.LOCATION_ID = Emp.Location_ID
Group By LOCATION_ID, STREET_ADDRESS, Employee_id
order by location_id;
For the second question:
SELECT FIRST_NAME,LAST_NAME,EMPLOYEE_ID,
count(Employee_id) as TIMES_HIRED
from HR.JOB_HISTORY inner join HR.EMPLOYEES On JOB_HISTORY.Employee_id = Employees.Employee_id
Group By FIRST_NAME,LAST_NAME,EMPLOYEE_ID
Having count(Employee_id) >1;