9. (16 points) Given $M(1,0)$, the line $y=t x+1$ intersects the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{8}=1$ at two distinct points $A$, $B$. Let the midpoint of segment $A B$ be $P$, find the range of the slope of line $P M$.
9. Solving the system of equations
$$
\left\{\begin{array}{l}
y=t x+1 \\
\frac{x^{2}}{4}-\frac{y^{2}}{8}=1
\end{array}\right.
$$

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, we get $-\frac{3}{2}<t<\frac{3}{2}$ and $t \neq \pm \sqrt{2}$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then
$$
x_{1}+x_{2}=\frac{2 t}{2-t^{2}}, \quad y_{1}+y_{2}=\left(t x_{1}+1\right)+\left(t x_{2}+1\right)=t\left(x_{1}+x_{2}\right)+2=\frac{4}{2-t^{2}} .
$$

The coordinates of the midpoint $P$ of segment $A B$ 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 line $P M$ exists, then $\frac{t}{2-t^{2}} \neq 1$, i.e., $t \neq 1$. Therefore,
$$
k=\frac{\frac{2}{2-t^{2}}}{\frac{t}{2-t^{2}}-1}=\frac{2}{t^{2}+t-2},
$$

and
$$
t^{2}+t-2=\left(t+\frac{1}{2}\right)^{2}-\frac{9}{4} \in\left[-\frac{9}{4}, 0\right) \cup(0, \sqrt{2}) \cup\left(\sqrt{2}, \frac{7}{4}\right),
$$

Thus, the range of $k=\frac{2}{t^{2}+t-2}$ is
$$
\left(-\infty,-\frac{8}{9}\right] \cup\left(\frac{8}{7}, \sqrt{2}\right) \cup(\sqrt{2},+\infty) \text {. }
$$