1. $\left|x^{2}-9\right|<\frac{1}{2} x^{2}+1$.
There are two intersection points: one is the intersection of $y=\frac{1}{2} x^{2}+1$ and $y=x^{2}-9$, and the other is the intersection of $y=\frac{1}{2} x^{2}+1$ and $y=9-x^{2}$.
$$
\begin{array}{r}
v=x^{2}-\theta=\frac{1}{2} x^{2}+1, \\
\frac{1}{2} x^{2}=10, x^{2}=20, \\
x= \pm 2 \sqrt{5} ;
\end{array}
$$

and
$$
\begin{array}{r}
9-x^{2}=\frac{1}{2} x^{2}+1, \\
8=\frac{3}{2} x^{2}, 16=3 x^{2}, \\
x= \pm \frac{4}{3} \sqrt{3} .
\end{array}
$$

As shown in Figure 11, all suitable $x$ values are contained in the set
$$
\left\{x \in R:-2 \sqrt{5}<x<-\frac{4}{3} \sqrt{3}\right.
$$

or $\left.\frac{4}{3} \sqrt{3}<x<2 \sqrt{5}\right\}$.