1. (Senior high school student work) Given the ellipse $C: \frac{(x-1)^{2}}{9}$ $+\frac{(y-2)^{2}}{4}=1$ on which there exist two points symmetric about the line $l_{1} y=2 x+m$, find the range of $n$.

1. Let $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$ be two points on $C$ that are symmetric with respect to $l$, and let $M(\bar{x}, \bar{y})$ be the midpoint of $A B$. Then
$$
\left\{\begin{array}{l}
4\left(x_{1}-1\right)^{2}+9\left(y_{1}-2\right)^{2}=36, \\
4\left(x_{2}-1\right)^{2}+9\left(y_{2}-2\right)^{2}=36, \\
\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \cdot 2=-1, \\
y=2 x+m .
\end{array}\right.
$$
(2) Subtracting (1) from (2) and simplifying gives $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=-\frac{4\left(x_{1}+x_{2}-2\right)}{9\left(y_{1}+y_{2}-4\right)}$.
(3) Substituting into (5) yields $9(\bar{y}-2)=8(\bar{x}-1)$.

From (4) and (B), we get $M\left(\frac{10-9 m}{10}, \frac{10-4 m}{5}\right)$.
Since $M$ is inside $C$, we have $\frac{\frac{\left(\frac{10-9 m}{10}-1\right)^{2}}{9+\left(\frac{10-4 m}{5}-2\right)^{2}}}{4}<1$.
Thus, $m^{2}<4$. Therefore, $-2<m<2$.