如果需要名称，请使用聚合和having子句：

select name, age
from mytable
where (course = 'Math' and score > 10) or
      (course = 'Lab' and score > 11) 
group by name, age
having count(distinct course) = 2;

如果需要详细记录，请使用窗口函数：

select t.*
from (select t.*,
             (dense_rank() over (partition by name, age order by course asc) +
              dense_rank() over (partition by name, age order by course desc)
             ) as cnt_unique_courses
      from mytable t
      where (course = 'Math' and score > 10) or
            (course = 'Lab' and score > 11) 
     ) t
where cnt_unique_courses = 2;

sql server不支持 count(distinct) 作为一个窗口函数。但是你可以用 dense_rank() 两次。

您可以查询同时满足这两个条件的记录—这是不可能的，因为每个记录都有一个课程。
您需要一个适用于具有相同名称的行的条件，因此建议改为聚合：

select age, name 
from mytable
where course in ('Math', 'Lab')
group by age, name 
having
    max(case when course = 'Math' then score end) > 10
    and max(case when course = 'Lab' then score end) > 11

如果您将问题表述为：
选择所有唯一的（姓名、年龄）组合
有一行分数>=10的课程数学
有一排分数大于11的课程实验室
然后您可以将其转换为sql中非常类似的内容：

select distinct t1.age, t1.name            -- unique combinations
from mytable t1
where exists ( select top 1 'x'            -- with a row math score >= 10
               from mytable t2
               where t2.name = t1.name
                 and t2.age = t1.age
                 and t2.course = 'math'
                 and t2.score >= 10 )
  and exists ( select top 1 'x'            -- with a row lab score > 11
               from mytable t3
               where t3.name = t1.name
                 and t3.age = t1.age
                 and t3.course = 'lab'
                 and t3.score > 11 );

我认为要么你的数据要么你的条件不适合你的输出。尽管基于您的条件，您可以单独使用您的条件，然后从两个选择中使用intersect并获取过滤后的数据。就像下面的代码。

select Age,Name
from Table_1
where Course ='Math' and Score>=10
INTERSECT
select Age,Name
from Table_1
where Course ='Lab' and Score>11

您可以使用共相关子查询编写查询

select * from table_1 t1
    where score >11 and course ='lab' 
        and [name] in (select [name] from table_1 t2 where t1.[name] =t2.[name] and t1.age =t2.Age
        and t2.Score >=10 and course = 'Math')