$94(-3,-2)$. The equation of line $l$ is
$$
y=k(x-1),(k>2) .
$$

Substituting into $y^{2}=4 x$ yields the equation
$$
k^{2} x^{2}-2\left(k^{2}+2\right) x+k^{2}=0 .
$$

Let the coordinates of point $M$ be $\left(x_{0}, y_{0}\right)$. Then
$$
\begin{array}{c}
x_{0}=\frac{x_{1}+x_{2}}{2}=\frac{k^{2}+2}{k^{2}}, \\
y_{0}=k\left(x_{0}-1\right)=\frac{2}{k} .
\end{array}
$$
The distance from $M$ to line $l_{m}$ is
$$
d=\frac{\left|3 \times \frac{k^{2}+2}{k^{2}}+4 \times \frac{2}{k}+m\right|}{5}=\frac{1}{5} .
$$

Thus,
$$
\frac{6}{k^{2}}+\frac{8}{k}+m+2=0 .(m>-3) .
$$

Let $\frac{1}{k}=t$, then
$$
m=-6 t^{2}-8 t-2,\left(0<-\frac{2}{3}\right) \text{ is a decreasing function, hence }
$$
$$
f\left(\frac{1}{2}\right)-3 \text{ implies } -3<m<-2.
$$
94 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 $\frac{1}{5}$ units away from the line $L_{m}: 3 x+4 y+m=0(m>-3)$, then the range of values for $m$ is . $\qquad$