1. Given
$$
\begin{aligned}
A= & \left\{x \mid x^{2}-4 x+3<0, x \in \mathbf{R}\right\}, \\
B= & \left\{x \mid 2^{1-x}+a \leqslant 0, x^{2}-2(a+7) x+5 \leqslant\right. \\
& 0, x \in \mathbf{R}\} .
\end{aligned}
$$

If $A \subseteq B$, then the range of real number $a$ is
$-1 .-4 \leqslant a \leqslant-1$.
From the problem, we know $A=\{x \mid 1<x<3\}$.
Let $f(x)=2^{1-x}+a$,
$$
g(x)=x^{2}-2(a+7) x+5 \text {. }
$$

To make $A \subseteq B$, it is only necessary that the graphs of $f(x)$ and $g(x)$ are below the x-axis in the interval $(1,3)$, then
$$
f(1) \leqslant 0, f(3) \leqslant 0, g(1) \leqslant 0, g(3) \leqslant 0 .
$$

From this, we solve $-4 \leqslant a \leqslant-1$.