1. Given the set $A=\left\{x \mid \log _{a}(a x-1)>1\right\}$. If $3 \in A$, then the range of values for $a$ is
$$
\text { -1. }\left(\frac{1}{3}, \frac{1}{2}\right) \cup(1,+\infty) \text {. }
$$

From the problem, we have
$$
\begin{array}{l}
\log _{a}(3 a-1)>1 \\
\Leftrightarrow\left\{\begin{array} { l } 
{ 0 < a < 1, } \\
{ 3 a-1 > a, } \\
{ \text{or} } \\
{ a > 1, } \\
{ 3 a-1 > a }
\end{array}\right.\right. \\
\Rightarrow a \in\left(\frac{1}{3}, \frac{1}{2}\right) \cup(1,+\infty) \text {. } \\
\end{array}
$$