4. Given that the line $l$ passes through the focus of the parabola $C: y^{2}=4 x$, and the slope $k>2, l$ intersects the parabola $C$ at points $A, B$; the midpoint $M$ of $A B$ is at a distance of $\frac{1}{5}$ from the line $L_{m}: 3 x+4 y+m=0(m>$ -3 ）. Then the range of values for $m$ is $\qquad$ .
4. $-32)$. Substituting this equation into $y^{2}=4 x$, eliminating $y$ yields
$$
k^{2} x^{2}-2\left(k^{2}+2\right) x+k^{2}=0 .
$$

Let $M\left(x_{0}, y_{0}\right)$, then
$$
x_{0}=\frac{x_{1}+x_{2}}{2}=\frac{k^{2}+2}{\dot{k}^{2}}, y=k\left(x_{0}-1\right)=\frac{2}{k} \text {. }
$$
The distance from $M$ to the line $L_{m}$ is
$$
\frac{\left|3 \cdot \frac{k^{2}+2}{k^{2}}+4 \cdot \frac{2}{k}+m\right|}{5}=\frac{1}{5} .
$$

Given $m>-3$, we have
$$
\frac{6}{k^{2}}+\frac{8}{k}+m+2=0 \text {. }
$$

Let $\frac{1}{k}=t$, then $0-\frac{2}{3}$ is a decreasing function, so
$$
f\left(\frac{1}{2}\right)-3$, then $-3<m<-2$.