select
  调查结果,
  sum(case when 年龄段 when 'man' then 1 else 0 end) as '中年人',
  sum(case when 年龄段 when 'boy' then 1 else 0 end) as '小孩',
  sum(case when 年龄段 when 'old' then 1 else 0 end) as '老年人',
  count(1) as 总计
from
  tb
group by
  调查结果

------解决方案--------------------

SQL code

declare @T table([调查结果] varchar(10),[man] int,[boy] int,[old] int)
insert @T
select '非常满意',10,50,45 union all
select '满意',45,45,12 union all
select '一般',21,21,21 union all
select '不满意',2,2,3 union all
select '非常不满意',0,0,0

select *,man+boy+old As 总计 from @T 
union all
select '总计',sum([man]),sum([boy]),
sum([old]),sum([man])+sum([boy])+sum([old]) from @T
/*
调查结果       man         boy         old         总计
---------- ----------- ----------- ----------- -----------
非常满意       10          50          45          105
满意         45          45          12          102
一般         21          21          21          63
不满意        2           2           3           7
非常不满意      0           0           0           0
总计         78          118         81          277
*/

------解决方案--------------------

SQL code

要后面的汇总可以用with 
with t as (
select
  调查结果,
  isnull(sum(case when 年龄类型 ='man' then 1 else 0 end),0) as '中年人',
  isnull(sum(case when 年龄类型 ='boy' then 1 else 0 end),0) as '小孩',
  isnull(sum(case when 年龄类型 ='old' then 1 else 0 end),0) as '老年人',
  count(1) as 总计
from
  tb a right join (
select 1 as 调查结果 
union all select 2 as 调查结果 
union all select 3 as 调查结果 
union all select 4 as 调查结果 
union all select 5 as 调查结果) b on b.调查结果=a.调查结果
group by
  b.调查结果
),
t1 as (
 select '总计' as 调查结果,sum(中年人) as 中年人,sum(小孩) as 小孩,
 sum(老年人) as 老年人, sum(中年人)+sum(小孩)+sum(老年人) as 总计 from t
),
t2 as (
 select '满意度' as 调查结果,sum(case when 调查结果!='不满意' and  调查结果!='非常不满意' then  
中年人 else 0 end)/sum(中年人) as 中年人,sum(case when 调查结果!='不满意' and 
 调查结果!='非常不满意' then  孩子 else 0 end)/sum(孩子) as 孩子,sum(case when
 调查结果!='不满意' and  调查结果!='非常不满意' then  
老年人 else 0 end)/sum(老年人) as 老年人，sum(case when
 调查结果!='不满意' and  调查结果!='非常不满意' then  
总计 else 0 end)/sum(总计) as 总计 from t 
),
t3 as (
 select '不满意度' as 调查结果,sum(case when 调查结果='不满意' or 调查结果='非常不满意' then  
中年人 else 0 end)/sum(中年人) as 中年人,sum(case when 调查结果='不满意' or 
 调查结果='非常不满意' then  孩子 else 0 end)/sum(孩子) as 孩子,sum(case when
 调查结果='不满意' or 调查结果='非常不满意' then  
老年人 else 0 end)/sum(老年人) as 老年人，sum(case when
 调查结果='不满意' or 调查结果='非常不满意' then  
总计 else 0 end)/sum(总计) as 总计 from t 
)
select * from t union all
select * from t1 union all
select * from t2 union all
select * from t3 union all

------解决方案--------------------
SQL code
create table test([id] int, [AgeLevel] varchar(10),[PROBLEM_ID] int)
go
insert test
select 1,'老年人', 1 union all
select 2,'小孩', 2 union all
select 3,'老年人', 3 union all
select 4,'中年人', 4 union all
select 5,'老年人', 5 union all
select 6,'老年人', 1 union all
select 7,'小孩', 2 union all
select 8,'小孩', 3 union all
select 9,'中年人', 4 union all
select 10,'中年人', 5 union all
select 11,'老年人',2

select * from test

declare @sql varchar(max)
declare @sql2 varchar(100)
declare @sql3 varchar(200)

select @sql=isnull(@sql+',','')+' count(case when [AgeLevel]='''+[AgeLevel]+''' then [PROBLEM_ID] end) a