设 $A=\left\{x \mid x^{2}+a x+12=0\right\}, B=\left\{x \mid x^{2}+3 x+2 b=0\right\}, A \cap B=\{2\}, C=\{2,-3\}$,

(1)求 $a, b$ 的值及 $A, B$;

(2)求 $(A \cup B) \cap C$.
【解】(1) $\because A \cap B=\{2\}, \therefore 4+2 a+12=0$, 即 $a=-8,4+6+2 b=0$, 即 $b=-5$,

$\therefore A=\left\{x \mid x^{2}-8 x+12=0\right\}=\{2,6\}, B=\left\{x \mid x^{2}+3 x-10=0\right\}=\{2,-5\}$.

(2) $\because A \cup B=\{-5,2,6\}, C=\{2,-3\}, \therefore(A \cup B) \cap C=\{2\}$.