8. Divide the elements of the set $M=\{1,2, \cdots, 12\}$ into three disjoint subsets: $M=A \cup B \cup C$, where,
\[
\begin{aligned}
A & =\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}, \\
B & =\left\{b_{1}, b_{2}, b_{3}, b_{4}\right\}, \\
C & =\left\{c_{1}, c_{2}, c_{3}, c_{4}\right\}\left(c_{1}<c_{2}<c_{3}<c_{4}\right),
\end{aligned}
\]

and $a_{k}+b_{k}=c_{k}(k=1,2,3,4)$.
Then the set $C$ is $\qquad$
8. $\{8,9,10,12\},\{7,9,11,12\},\{6,10,11,12\}$.

From $\sum_{i=1}^{4} c_{i}=\sum_{i=1}^{4} a_{i}+\sum_{i=1}^{4} b_{i}$, we get
$2 \sum_{i=1}^{4} c_{i}=\sum_{i=1}^{4} a_{i}+\sum_{i=1}^{4} b_{i}+\sum_{i=1}^{4} c_{i}$
$=1+2+\cdots+12=78$.
Therefore, $\sum_{i=1}^{4} c_{i}=39$.
Without considering the pairing scenarios, let $c_{1}27,10 \leqslant c_{3} \leqslant 11$, and $c_{2}8$.
Thus, $c_{2}$ can only be 9 or 10, and $c_{1}$ can only be 7 or 6.
Hence, $C=\{7,9,11,12\}, C=\{6,10,11,12\}$.
On the other hand, all three scenarios correspond to the respective subsets $A$ and $B$ (as shown in figures 7, 8, 9).