13. Given the sets $A=\left\{(x, y) \left\lvert\, \frac{y-3}{x-2}=a+1\right.\right\}, B=$ $\left\{(x, y) \mid\left(a^{2}-1\right) x+(a-1) y=15\right\}$. Then the possible values of $a$ that make $A \cap B=$ $\varnothing$ are . $\qquad$
13. $-4,-1,1, \frac{5}{2}$.
$A$ is $\{(x, y) \mid(a+1) x-y-2 a+1=0, x \neq 2\}$.
$A \cap B=\varnothing$ occurs only in one of the following three cases:
(1) When $a=-1$, since $A=\{(x, y) \mid y=3, x \neq 2\}$, $B=\left\{(x, y) \left\lvert\, y=-\frac{15}{2}\right.\right\}$, the lines $y=3$ and $y=-\frac{15}{2}$ are parallel, so $A \cap B=\varnothing$;
(2) When $a \neq-1$, and $\frac{a^{2}-1}{a+1}=\frac{a-1}{-1} \neq \frac{-15}{-2 a+1}$, the line $(a+1) x-y-2 a+1=0$ is parallel to $\left(a^{2}-1\right) x+(a-1) y=15$, so $A \cap B=\varnothing$. In this case, $a=1$.
(3) When $a \neq \pm 1$, and $(2,3) \in B$, the line $(a+1) x-y-2 a+1=0$ intersects with $\left(a^{2}-1\right) x+(a-1) y=15$ at the point $(2,3)$, but $(2,3) \notin A$, so $A \cap B=\varnothing$. In this case, $\left(a^{2}-1\right) \times 2+(a-1) \times 3=15$, solving for $a$ gives $a=-4$ or $a=\frac{5}{2}$.