13. Given the set $A=\left\{t \mid t\right.$, such that $\left.\left\{x \mid x^{2}+2 t x-4 t-3 \neq 0\right\}=\mathbf{R}\right\}, B=\{t \mid t$, such that $\left.\left\{x \mid x^{2}+2 t x-2 t=0\right\} \neq \varnothing\right\}$, where $x, t$ are real numbers, find $A \cap B$.
13. From the problem, we know $\left\{x \mid x^{2}+2 t x-4 t-3 \neq 0\right\}=\mathbf{R}$, so $\Delta_{1}=4 t^{2}-4(-4 t-3)<0$, which gives $-3 < t < -1$, i.e., $A=(-3,-1)$. Also, $\left\{x \mid x^{2}+2 t x-2 t=0\right\} \neq \varnothing$, so $\Delta_{2}=4 t^{2}+8 t \geqslant 0$, which gives $t \leqslant -2$ or $t \geqslant 0$, i.e., $B=(-\infty,-2] \cup[0,+\infty)$. Therefore, $A \cap B=(-3,-2]$.