The intercepts on the straight $y=mx$ by the line $y=2$ and $y=6$ is less than $5$, then $m$ belongs to

$\begin{array}{l} y=m x \quad \& \quad y=2 \\ 2=m x \quad A(2 / m, 2) \\ x=2 / m \\ y=m x & y=6 \\ (3 m-4)(3 m+4)>0 \\ \frac{t+\frac{1}{-4 / 3}-\frac{1}{3}}{1} \\ m \in(-\infty,-4 / 3) \cup(4 / 3, \infty) \\ \end{array}$