13. If $0<a<\sqrt{3} \cos \theta, \theta \in\left[-\frac{\pi}{4}, \frac{\pi}{3}\right]$, find the minimum value of $f(a, \theta)=\cos ^{3} \theta+\frac{4}{3 a \cos ^{2} \theta-a^{3}}$.
13. $\frac{17 \sqrt{2}}{4}$

Since $00$.
$$
\dot{a}^{2}\left(3 \cos ^{2} \theta-a^{2}\right)^{2} \leqslant \frac{1}{2}\left[\frac{2 a^{2}+\left(3 \cos ^{2} \theta-a^{2}\right)+\left(3 \cos ^{2} \theta-a^{2}\right)}{3}\right]^{3}=4 \cos ^{6} \theta
$$

Equality holds if and only if $2 a^{2}=3 \cos ^{2} \theta-a^{2}$, i.e., $a=\cos \theta$.
$$
f(a, \theta)=\cos ^{3} \theta+\frac{4}{3 a \cos ^{2} \theta-a^{3}} \geqslant \cos ^{3} \theta+\frac{2}{\cos ^{3} \theta}
$$

Let $\cos ^{3} \theta=t\left(\frac{1}{8} \leqslant t \leqslant \frac{\sqrt{2}}{4}\right), h(t)=t+\frac{2}{t} \geqslant 2 \sqrt{2}$.
Equality holds if and only if $t=\sqrt{2}$.
It can be proven that: $h(t)$ is monotonically decreasing in $\left[\frac{1}{8}, \frac{\sqrt{2}}{4}\right]$. Therefore, $f(a, \theta)_{\operatorname{min}}=\frac{17 \sqrt{2}}{4}$.