$$
\begin{array}{l}
\text { 1. Let } A=\left\{x \mid x^{2}<\log _{a} x\right\} , \\
B=\left\{y \mid y=2^{x}, x \in(-\infty,-1)\right\} \text {. }
\end{array}
$$

If $A \varsubsetneqq B$, then the range of values for $a$ is
$-1 \cdot\left(0, \frac{1}{16}\right) \cup\left[\mathrm{e}^{\frac{1}{2 \infty}},+\infty\right)$.
When $0\log _{a} \frac{1}{2} \Rightarrow a \in\left(0, \frac{1}{16}\right)$.
When $a>1$, $A \neq \varnothing, x^{2} \geqslant \log _{a} x$.
Assume the function $y=x^{2}$ and $y=\log _{a} x$ are tangent at the point $\left(x_{0}, y_{0}\right)$, and their common tangent line passes through the point $\left(x_{0}, y_{0}\right)$. Then
$$
\left\{\begin{array}{l}
2 x_{0}=\frac{1}{x_{0} \ln a}, \\
x_{0}^{2}=\log _{a} x_{0}
\end{array} \Rightarrow x_{0}=\mathrm{e}^{\frac{1}{2}}, a=\mathrm{e}^{\frac{1}{2 c}} .\right.
$$

Combining with the graph, we know $a \geqslant \mathrm{e}^{\frac{1}{2 c}}$.
In summary, $a \in\left(0, \frac{1}{16}\right) \cup\left[\mathrm{e}^{\frac{1}{2}},+\infty\right)$.