1's right focus coincides, if the line $y=t x+1$ intersects the hyperbola $C$ at two distinct points $A$ and $B$, let the midpoint of $A$ and $B$ be $P$, find the range of the slope $k$ of the line $P M$.

According to the problem, the coordinates of the focus of the parabola are $(2 \sqrt{3}, 0)$, then $b^{2}=12-4=8$, so the equation of the hyperbola $C$ is $\frac{x^{2}}{4}-\frac{y^{2}}{8}=1$.

Solving the system of equations $\left\{\begin{array}{l}y=t x+1, \\ 2 x^{2}-y^{2}=8,\end{array}\right.$ we get $\left(2-t^{2}\right) x^{2}-2 t x-9=0$, then $\left\{\begin{array}{l}2-t^{2} \neq 0, \\ \Delta=4 t^{2}+36\left(2-t^{2}\right)=8\left(9-4 t^{2}\right)>0,\end{array}\right.$ solving this yields $-\frac{3}{2}<t<\frac{3}{2}$ and $t \neq \pm \sqrt{2}$.

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then the coordinates of the midpoint $P$ of $A B$ are $\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$, i.e., $\left(\frac{t}{2-t^{2}}, \frac{2}{2-t^{2}}\right)$. Since the slope $k$ of the line $P M$ exists, then $\frac{t}{2-t^{2}} \neq 1$, i.e., $t \neq 1$. Therefore, $k=$ $\frac{\frac{2}{\frac{t}{t}}}{\frac{t}{2-t^{2}}-1}=\frac{2}{t^{2}+t-2}$, and $t^{2}+t-2=\left(t+\frac{1}{2}\right)^{2}-\frac{9}{4} \in\left[-\frac{9}{4}, 0\right) \cup(0, \sqrt{2}) \cup$ $\left(\sqrt{2}, \frac{7}{4}\right)$, so the range of $k=\frac{2}{t^{2}+t-2}$ is $\left(-\infty,-\frac{8}{9}\right] \cup\left(\frac{8}{7}, \sqrt{2}\right) \cup(\sqrt{2}$, $+\infty)$.