我的解题内容：

with t as(
select a.emp_id,a.exam_id,b.emp_level,case when (a.exam_id=9001 and score>78.667 and round(timestampdiff(second,start_time,submit_time)/60,1) < 27.3) then a.emp_id else null end 9001isgood,
case when (a.exam_id=9002 and score>75.833 and round(timestampdiff(second,start_time,submit_time)/60,1) < 34.2) then a.emp_id else null end 9002isgood
from   exam_record a 
left join emp_info  b on a.emp_id  =b.emp_id
where b.emp_level < 7 )

select t.emp_id,emp_level,c.tag
from t
left join examination_info c on c.exam_id=t.exam_id
where 9001isgood is not null or 9002isgood is not null

运用了窗口函数的便捷：通过窗口函数动态计算 每个考试（所有 exam_id） 的平均分数和平均耗时，然后通用化判断员工在对应考试中是否 “分数高于该考试平均分、耗时低于该考试平均耗时”，且员工级别 < 7。

with t as (select *,
round(avg(score) over (partition by exam_id),3) as examavgscore,
round(avg(timestampdiff(second,start_time,submit_time)/60) over (partition by exam_id),1) as avgtime
from exam_record )

select t.emp_id,emp_level,tag
from t 
left join emp_info a on a.emp_id=t.emp_id
left join examination_info b on  b.exam_id=t.exam_id
where emp_level<7 and  score>examavgscore and round(timestampdiff(second,start_time,submit_time)/60,1)<avgtime
order by t.emp_id