8. 椭圆 $\frac{x^{2}}{9}+\frac{y^{2}}{2}=1$ 的焦点为 $F_{1}, F_{2}$, 点 $P$ 在粗圆上. 若 $\left|P F_{1}\right|=4$,则 $\left|P F_{2}\right|=$ \$ \qquad \$ ; $\angle F_{1} P F_{2}$ 的大小为 \$ \qquad \$
： $\because\left|P F_{1}\right|+\left|P F_{2}\right|=2 \times 3=6$, 且 $\left|P F_{1}\right|=4$,

$\therefore\left|P F_{2}\right|=2$.

在 $\triangle F_{1} P F_{2}$ 中, $\left|P F_{1}\right|=4,\left|P F_{2}\right|=2,\left|F_{1} F_{2}\right|=2 \sqrt{7}$,

$\therefore \cos \angle F_{1} P F_{2}=\frac{\left|P F_{1}\right|^{2}+\left|P F_{2}\right|^{2}-\left|F_{1} F_{2}\right|^{2}}{2\left|P F_{1}\right| \cdot\left|P F_{2}\right|}$

$=-\frac{1}{2}, \therefore \angle F_{1} P F_{2}=120^{\circ}$.

答案： $2120^{\circ}$