8. Given a hyperbola $C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with an eccentricity of $\frac{\sqrt{6}}{2}$, the left and right foci are $F_{1}, F_{2}$, and $P$ is a point on the hyperbola. Let $R, r$ be the circumradius and inradius of $\triangle P F_{1} F_{2}$. If $\angle F_{1} P F_{2}=60^{\circ}$, then $\frac{R}{r}=$ $\qquad$ .
Given
$e=\frac{\sqrt{6}}{2} \Rightarrow \frac{b^{2}}{a^{2}}=e^{2}-1=\frac{1}{2}$, let's assume $a=\sqrt{2}, b^{2}=1, c=\sqrt{3}$, then $2 R=\frac{2 \sqrt{3}}{\sin 60^{\circ}}$
$\Rightarrow R=2$. Let $\left|P F_{1}\right|=x,\left|P F_{2}\right|=y$, thus $\left\{\begin{array}{l}12=x^{2}+y^{2}-x y, \\ x-y=2 \sqrt{2}\end{array}\right.$
$\Rightarrow x+y=2 \sqrt{6}$. Also, $S_{\triangle P F_{1} F_{2}}=b^{2} \cot 30^{\circ}=\sqrt{3}$, hence $r=\frac{2 S_{\triangle P F_{1} F_{2}}}{x+y+2 c}$
$=\frac{\sqrt{3}}{\sqrt{6}+\sqrt{3}}=\sqrt{2}-1$. Therefore, $\frac{R}{r}=\frac{2}{\sqrt{2}-1}=2 \sqrt{2}+2$.