6. Let the ellipse $C: \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$, and the line $l: y=4x+m$. Then, the range of values for $m$ such that there are two points on the ellipse $C$ that are symmetric with respect to the line $l$ is $\qquad$.
6. $\left(-\frac{2 \sqrt{13}}{13}, \frac{2 \sqrt{13}}{13}\right)$

Analysis: Let the two points on the ellipse symmetric about the line $l$ be $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, and the midpoint of $A B$ be $M\left(x_{0}, y_{0}\right)$.
From $\left\{\begin{array}{l}\frac{x_{1}^{2}}{4}+\frac{y_{1}^{2}}{3}=1 \\ \frac{x_{2}^{2}}{4}+\frac{y_{2}^{2}}{3}=1\end{array}\right.$, subtracting the two equations gives $\frac{x_{1}^{2}-x_{2}^{2}}{4}+\frac{y_{1}^{2}-y_{2}^{2}}{3}=0$, which simplifies to
$\frac{\left(x_{1}+x_{2}\right)\left(x_{1}-x_{2}\right)}{4}+\frac{\left(y_{1}+y_{2}\right)\left(y_{1}-y_{2}\right)}{3}=0$. Using the slope of line $A B$, $k_{A B}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$, and the midpoint formula $x_{0}=\frac{x_{1}+x_{2}}{2}, y_{0}=\frac{y_{1}+y_{2}}{2}$, we get $\frac{x_{0}}{4}+\frac{y_{0}}{3} \cdot k_{A B}=0$, which simplifies to $k_{A B}=-\frac{3 x_{0}}{4 y_{0}}$. Since $l \perp A B$, we have
$k_{A B}=-\frac{1}{4}$, so $-\frac{1}{4}=-\frac{3 x_{0}}{4 y_{0}}$. Also, point $M$ lies on $l$, so $y_{0}=4 x_{0}+m$, solving this gives $M(-m,-3 m)$. According to the condition that $M$ is inside the ellipse, we have $\frac{(-m)^{2}}{4}+\frac{(-3 m)^{2}}{3}<1$, thus $-\frac{2 \sqrt{13}}{13}<m<\frac{2 \sqrt{13}}{13}$.