1. Let set $A=\left\{x \mid x^{2}-3 x-10 \leq 0\right\}, B=\{x \mid m+1 \leq x \leq 2 m-1\}$, if $A \cap B=B$, then the range of real number $m$ is $\qquad$.
Given $m \leq 3$.
Analysis From $A \cap B=B$, we know that $B \subseteq A$, and $A=\left\{x \mid x^{2}-3 x-10 \leq 0\right\}=\{x \mid-2 \leq x \leq 5\}$.
When $B=\varnothing$, $m+1>2 m-1$, i.e., $m<2$, in this case $B \subseteq A$ holds;
When $B \neq \varnothing$, $m+1 \leq 2 m-1$, i.e., $m \geq 2$, by $B \subseteq A$, we get $\left\{\begin{array}{l}-2 \leq m+1 \\ 2 m-1 \leq 5\end{array}\right.$, solving this yields $-3 \leq m \leq 3$, and since $m \geq 2$, we have $2 \leq m \leq 3$.
In summary, we have $m \leq 3$.