[Solution] Suppose when $m+10$ is replaced by $m$, the average value increases by 2, then we know the sequence has $10 \div 2=5$ terms. Also, when $m-8$ is replaced by $m$, the median is $m-4$.
Arranging the first three terms in ascending order should be: $10, m-4, m$.
Since 32 is the mode, clearly at least two terms are 32, clearly $m \neq 32$, otherwise the average value of the sequence would be greater than 22. Let the five numbers be $10, m-4, m, 32, 32$, then we have
$$
10+(m-4)+m+32+32=22 \times 5,
$$

Solving for $m$ gives $m=20$.
Therefore, the answer is $(E)$.
28.24 The mode of a list of integers is 32, the average is 22, the smallest number is 10, and the median is $m$. If $m+10$ replaces $m$, the new average is 24, and the new median is $m+10$; if $m-8$ replaces $m$, the new median is $m-4$. What is the value of $m$?
(A) 16.
(B) 17.
(C) 18.
(D) 19.
(E) 20.
(48th American High School Mathematics Examination, 1997)