drop table if exists heyf_t10;
create table heyf_t10 (empid int ,deptid int ,salary decimal(10,2) );
insert into heyf_t10 values
    (1,10,5500.00),
    (2,10,4500.00),
    (3,20,1900.00),
    (4,20,4800.00),
    (5,40,6500.00),
    (6,40,14500.00),
    (7,40,44500.00),
    (8,50,6500.00),
    (9,50,7500.00);

select t.* 
from (
    select distinct a.empid eid, a.* 
    from heyf_t10 a 
    order by salary desc 
) t 
group by t.deptid

eid    empid    deptid    salary
1      1        10        5500.00
4      4        20        4800.00
7      7        40        44500.00
9      9        50        7500.00

select a.*
from heyf_t10 a, heyf_t10 b
where a.deptid = b.deptid
and a.salary <= b.salary
group BY a.deptid, a.salary
having count(a.deptid) <= 2
order by a.deptid, a.salary desc

empid    deptid    salary
1        10        5500.00
2        10        4500.00
4        20        4800.00
3        20        1900.00
7        40        44500.00
6        40        14500.00
9        50        7500.00
8        50        6500.00

select deptid
    , substring_index(group_concat(salary order by salary desc), ',', 2) salarTop2
    , group_concat(salary order by salary desc) salarys
from heyf_t10
group by deptid

deptid    salarTop2          salarys
10        5500.00,4500.00    5500.00,4500.00
20        4800.00,1900.00    4800.00,1900.00
40        44500.00,14500.00  44500.00,14500.00,6500.00
50        7500.00,6500.00    7500.00,6500.00