1. Given non-empty set $A=\{x \mid m+1 \leqslant x \leqslant 2 m-1\}, B=\left\{x \mid x^{2}-2 x-15 \leqslant 0\right\}$, and $A \subseteq B$, then the range of real number $m$ is $\qquad$ .
Solve
$x^{2}-2 x-15=(x+3)(x-5) \leqslant 0 \Rightarrow B=[-3,5]$, thus $\left\{\begin{array}{l}m+1 \leqslant 2 m-1, \\ m+1 \geqslant-3, \\ 2 m-1 \leqslant 5\end{array} \Rightarrow 2 \leqslant m \leqslant 3\right.$, so the range of real number $m$ is $[2,3]$.