5. Let the complex numbers $z_{1}, z_{2}$ satisfy $\left|z_{1}\right|=\left|z_{1}+z_{2}\right|=3,\left|z_{1}-z_{2}\right|=3 \sqrt{3}$, then $\log _{3}\left|\left(z_{1} \overline{z_{2}}\right)^{2000}+\left(\overline{z_{1}} z_{2}\right)^{2000}\right|=$ $\qquad$ .
4000
5. 【Analysis and Solution】Solution 1 From the given conditions, we have
$$
\begin{array}{l}
9=\left|z_{1}+z_{2}\right|^{2}=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}, \\
27=\left|z_{1}-z_{2}\right|^{2}=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-\left(z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}\right) .
\end{array}
$$

Since $\left|z_{1}\right|=3$, it follows that $\left|z_{2}\right|=3, z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}=-9$, and $\left|z_{1} \bar{z}_{2}\right|+| \bar{z}_{1} z_{2} \mid=9$.
Let $z_{1} \bar{z}_{2}=9(\cos \theta+i \sin \theta)$,
$$
\bar{z}_{1} z_{2}=9(\cos \theta-i \sin \theta) \text {. }
$$
$-9=z_{1} \bar{z}_{2}+z_{1} \bar{z}_{2}=18 \cos \theta$, which gives
$$
\cos \theta=-\frac{1}{2} \text {. }
$$

Thus, $z_{1} \bar{z}_{2}=9 \omega$ or $z_{1} \bar{z}_{2}=9 \omega^{2}$,
where $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} i$.
When $z_{1} \bar{z}_{2}=9 \omega$, we get
$$
\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}=-9^{2000} \text {, }
$$

Therefore, $\log _{3}\left|\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}\right|=4000$.
When $z_{1} \bar{z}_{2}=9 \omega^{2}$, we get the same result, so the answer is 4000.
Solution 2 Consider the geometric meaning of complex numbers. Let the points corresponding to the complex numbers $z_{1}, z_{2}$ in the complex plane be $A, B$, and let $C$ be the midpoint of $A B$. From the conditions,
$$
\begin{array}{l}
|O A|=\left|z_{1}\right|=3, \\
|O C|=\frac{1}{2}\left|z_{1}+z_{2}\right|=\frac{3}{2},
\end{array}
$$
$|A C|=\frac{1}{2}\left|z_{1}-z_{2}\right|=\frac{3}{2} \sqrt{3}$, thus it is easy to see that $\triangle O A B$ is an equilateral triangle. Without loss of generality, assume $z_{2}=z_{1} e^{i \frac{\pi}{3}}$, substituting this in gives $\log _{3}\left|\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}\right|=4000$.