(12) Given that point $P$ is a moving point on the hyperbola $\frac{x^{2}}{8}-\frac{y^{2}}{4}=1$, $F_{1}$ and $F_{2}$ are its left and right foci respectively, and $O$ is the origin, then the range of $\frac{\left|P F_{1}\right|+\left|P F_{2}\right|}{|O P|}$ is $\qquad$
(12) $(2, \sqrt{6}]$ Hint: By the symmetry of the hyperbola, we may assume $P(x, y)(x \geqslant 2 \sqrt{2})$.
By the focal radius formula, we have
$$
\begin{array}{l}
\left|P F_{1}\right|=\frac{\sqrt{6}}{2} x+2 \sqrt{2}, \\
\left|P F_{2}\right|=\frac{\sqrt{6}}{2} x-2 \sqrt{2},
\end{array}
$$

Therefore,
$$
|O P|=\sqrt{x^{2}+y^{2}}=\sqrt{x^{2}+\frac{x^{2}}{2}-4}=\sqrt{\frac{3}{2} x^{2}-4},
$$

Thus,
$$
\frac{\left|P F_{1}\right|+\left|P F_{2}\right|}{|O P|}=\frac{\sqrt{6} x}{\sqrt{\frac{3}{2} x^{2}-4}}=\frac{\sqrt{6}}{\sqrt{\frac{3}{2}-\frac{4}{x^{2}}}} .
$$

Given $x \geqslant 2 \sqrt{2}$, hence
$$
\begin{aligned}
1 & \leqslant \sqrt{\frac{3}{2}-\frac{4}{x^{2}}}<\frac{\sqrt{6}}{2}, \\
& \Rightarrow \frac{2}{\sqrt{6}}<\frac{1}{\sqrt{\frac{3}{2}-\frac{4}{x^{2}}}} \leqslant 1, \\
& \Rightarrow 2<\frac{\sqrt{6}}{\sqrt{\frac{3}{2}-\frac{4}{x^{2}}}} \leqslant \sqrt{6} .
\end{aligned}
$$