1. Let the set
$$
A=\left\{5, \log _{2}(a+3)\right\}, B=\{a, b\}(a, b \in \mathbf{R}) \text {. }
$$

If $A \cap B=\{1\}$, then $A \cup B=$ $\qquad$
$-、 1 .\{5,1,-1\}$.
Since $A \cap B=\{1\}$, therefore, $\log _{2}(a+3)=1$
Solving for $a$ gives $a=-1$. Thus, $b=1$.
Hence, $A \cup B=\{5,1,-1\}$.