12. Given that $P$ 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, 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. $(2, \sqrt{6}]$.

By the symmetry of the hyperbola, we can assume $P(x, y)$ $(x \geqslant 2 \sqrt{2})$.
By the focal radius formula, we get
$$
\begin{array}{l}
\left|P F_{1}\right|=\frac{\sqrt{6}}{2} x+2 \sqrt{2}, \\
\left|P F_{2}\right|=\frac{\sqrt{6}}{2} x-2 \sqrt{2}, \\
|O P|=\sqrt{x^{2}+y^{2}}=\sqrt{x^{2}+\frac{x^{2}}{2}-4} \\
=\sqrt{\frac{3}{2} x^{2}-4} .
\end{array}
$$

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