9. Given $\left\{A=|x| x^{2}-4 x+3<0, x \in \mathbf{R}\right\}, B=\left\{x \mid 2^{1-x}+a \leqslant 0, x^{2}-2(a+7) x+5 \leqslant 0, x \in\right.$ $\mathbf{R}\}$. If $A \subseteq B$, then the range of real number $a$ is $\qquad$
$$
-4 \leqslant a \leqslant-1
$$
9. 【Analysis and Solution】It is easy to get $A=(1,3)$, let $f(x)=2^{1-x}+a, g(x)=x^{2}-2(a+7) x+5$.

To make $A \subseteq B$, it is only necessary for the graphs of $f(x)$ and $g(x)$ to be below the x-axis on $(1,3)$. The necessary and sufficient condition for this is: simultaneously, $f(1) \leqslant 0, f(3) \leqslant 0, g(1) \leqslant 0, g(3) \leqslant 0$. This leads to $-4 \leqslant a \leqslant-1$.