8. Let the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ have an eccentricity $e=\frac{\sqrt{3}}{2}$, and its latus rectum (the focal diameter perpendicular to the major axis) $d=1, F_{1} 、 F_{2}$ be the two foci, $P$ be any point on the ellipse except the endpoints of the major axis, and the angle bisector $P M$ of $\angle F_{1} P F_{2}$ intersect the major axis at point $M(m, 0)$. Then the range of $m$ is . $\qquad$
8. $\left(-\frac{3}{2}, \frac{3}{2}\right)$.

Substitute $x=c$ into the equation to get the latus rectum
$d=\frac{2 b^{2}}{a}=1$
$\Rightarrow a=2 b^{2}$.
From $e=\frac{c}{a}=\sqrt{\frac{a^{2}-b^{2}}{a^{2}}}=\frac{\sqrt{3}}{2}$
$\Rightarrow a=2 b$.
Solving equations (1) and (2) simultaneously, we get
$a=2, b=1$.
Thus, $c=\sqrt{3}$.
Therefore, the ellipse $E: \frac{x^{2}}{4}+y^{2}=1$.
From equation (3), the equation of the tangent line at point $P\left(x_{0}, y_{0}\right)$ is
$$
\frac{x_{0} x}{4}+y_{0} y=1 \text {, }
$$

The slope is $k=-\frac{x_{0}}{4 y_{0}}$.
According to the tangent line theorem of the ellipse, the normal line $P M$ at point $P\left(x_{0}, y_{0}\right)$ has a slope of $k^{\prime}=-\frac{1}{k}=\frac{4 y_{0}}{x_{0}}$.
Thus, $l_{P M}: y-y_{0}=\frac{4 y_{0}}{x_{0}}\left(x-x_{0}\right)$.
Let $y=0$, we get $m=\frac{3 x_{0}}{4}$.
Given that $x_{0} \in(-a, a)$.
Hence, $m \in\left(-\frac{3}{2}, \frac{3}{2}\right)$.