2. Given $A=\left\{x \mid x^{2}-4 x+3<0, x \in \mathbf{R}\right\}, B=\left\{x \mid 2^{1-x}+a \leqslant 0\right.$ and $x^{2}-2(a+7) x+5 \leqslant 0$, $x \in \mathbf{R}\}$, if $A \leqslant B$, then the range of real number $a$ is $\qquad$
2. $-4 \leqslant a \leqslant-1$. It is easy to get $A=\{x \mid 1<x<3\}$. 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 that the graphs of $f(x), g(x)$ are below the x-axis within $(1,3)$. The necessary and sufficient condition for this is $f(1) \leqslant 0, f(3) \leqslant 0, g(1) \leqslant 0, g(3) \leqslant$ 0, from which we can deduce $-4 \leqslant a \leqslant-1$.