12. Given $A=\left\{x \mid x^{2}-4 x+3<0, x \in \mathbf{R}\right\}, B=\left\{x \mid 2^{1}+a \leqslant 0, x^{2}-2(a+7) x+5\right.$ $\leqslant 0, x \in \mathbf{R}\}$, if $A \subseteq B$, then the range of real number $a$ is $\qquad$
12. $-4 \leqslant a \leqslant-1$. Let $C=\left\{x \mid 2^{1-x}+a \leqslant 0, x \in \mathbf{R}\right\}, D=\left\{x \mid x^{2}-2(a+7) x+5 \leqslant 0, x \in \mathbf{R}\right\}$. Then $B=C \cap D$. So $A \subseteq B$ implies: $A \subseteq C$ and $A \subseteq D$. And $A=\{x \mid 1<x<3\}$, so the graphs of $f(x)=2^{1-x}+a \cdot g(x)$ and $g(x)=x^{2}-2(a+7) x+5$ are both below the x-axis in the interval (1,3).
Therefore, $\left\{\begin{array}{l}f(1) \leqslant 0 \\ f(3) \leqslant 0 \\ g(1) \leqslant 0 \\ g(3) \leqslant 0\end{array} \Rightarrow-4 \leqslant a \leqslant-1\right.$.