13. $O$ is the pole of the polar coordinate system, the hyperbola $\rho=\frac{a^{2}}{5-6 \sqrt{2} \cos \theta}$ intersects the line $\rho \cos \left(\theta-\frac{\pi}{4}\right)=$ $a$ at points $A, B$. Find the range of values for $a$ such that $\angle A O B \leqslant \frac{2 \pi}{3},(\rho$ can take negative values $)$.
13. According to the problem: $\frac{a^{2}}{5-6 \sqrt{2} \cos \theta} \cos \left(\theta-\frac{\pi}{4}\right)=a$
$\Leftrightarrow \frac{a \sqrt{2}}{2} \cos \theta+\frac{a^{2} \sqrt{2}}{2} \sin \theta=5 a-6 \sqrt{2} a \cos \theta$
$\Leftrightarrow\left(\frac{a^{2} \sqrt{2}}{2}+6 \sqrt{2} a\right) \cos \theta+\frac{a^{2} \sqrt{2}}{2} \sin \theta=5 a$
$\Leftrightarrow \sqrt{\left(\frac{a \sqrt{2}}{2}+6 \sqrt{2}\right)^{2}+\left(\frac{a \sqrt{2}}{2}\right)^{2}} \cos (\theta-\varphi)=5$
where $\varphi=\operatorname{arccot} \frac{\frac{a^{2} \sqrt{2}}{2}}{\frac{a^{2} \sqrt{2}}{2}+6 \sqrt{2} a}$
According to the problem, the equation has two solutions $\theta_{1}, \theta_{2}\left(0 \leqslant \theta_{1} \leqslant \theta_{2}<2 \pi\right)$ and $\theta_{2}-\theta_{1} \leqslant \frac{2 \pi}{3}$
or $\theta_{2}-\theta_{1} \geqslant \frac{4 \pi}{3}$, let $2 \alpha=\theta_{2}-\theta_{1}$
Also, $\cos \left(\theta_{1}-\varphi\right)=\cos \left(\theta_{2}-\varphi\right)$
$\Rightarrow \theta_{1}-\varphi=\theta_{2}-\varphi+2 k \pi$
or $\theta_{1}-\varphi+\theta_{2}-\varphi=2 k \pi(k \in \mathbf{N})$

Also, $0 \leqslant \theta_{1}<\theta_{2}<2 \pi$
So $2\left(\theta_{1}-\varphi\right)+2 \alpha=2 k \pi$
$$
\theta_{1}-\varphi=k \pi-\alpha
$$

Also, $0<\alpha \leqslant \frac{\pi}{3}$ or $\frac{2 \pi}{3}<\alpha<\pi$
So $\cos \left(\theta_{1}-\varphi\right) \in\left(-1,-\frac{1}{2}\right] \cup\left[\frac{1}{2}, 1\right)$
So $\frac{5}{\sqrt{\left(\frac{a \sqrt{2}}{2}+6 \sqrt{2}\right)^{2}+\left(\frac{a \sqrt{2}}{2}\right)^{2}}} \in\left[\frac{1}{2}, 1\right)$
So $25<\frac{a^{2}}{2}+12 a+72+\frac{a^{2}}{2} \leqslant 100-11<(a+6)^{2} \leqslant 64=8^{2}$ $\Leftrightarrow-2 \leqslant a \leqslant 14$
where $a=0$ does not satisfy the problem.
So the range of $a$ is $[-2,0) \cup(0,14]$.