8.  Let $A$ be a 20-element subset of the set
$$
M=\{1,2, \cdots, 2012\}
$$

such that the difference between any two elements in $A$ is a multiple of 12. Then the number of such subsets $A$ is $\qquad$
8. $8 C_{168}^{20}+4 C_{167}^{20}$.

For $x, y \in M$, if $12 \mid (x-y)$, then $x, y$ are called "congruent".

Thus, when $n \in \{1, 2, \cdots, 8\}$, $n$ has 168 congruent numbers. For each such $n$, the 168-element set
$$
T_{n}=\{n+12 k \mid k=0,1, \cdots, 167\}
$$

has any 20-element subset that meets the condition, yielding a total of $8 C_{168}^{20}$ subsets;

When $n \in \{9, 10, 11, 12\}$, $n$ has 167 congruent numbers. For each such $n$, the 167-element set
$$
T_{n}=\{n+12 k \mid k=0,1, \cdots, 166\}
$$

has any 20-element subset that also meets the condition, yielding a total of $4 C_{167}^{20}$ subsets.

Therefore, the number of required sets is $8 C_{168}^{20}+4 C_{167}^{20}$.