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
$$
\log _{2}\left|\left(z_{1} \overline{z_{2}}\right)^{2000}+\left(\overline{z_{1}} z_{2}\right)^{2000}\right|=
$$
5. From the given conditions, we have
$$
\begin{array}{c}
9=\left|z_{1}+z_{2}\right|^{2}=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+z_{1} \overline{z_{2}}+\overline{z_{1}} z_{2}, \\
27=\left|z_{1}-z_{2}\right|^{2}=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-\left(z_{1} \overline{z_{2}}+\overline{z_{1}} z_{2}\right) .
\end{array}
$$

Since $\left|z_{1}\right|=3$, it follows that $\left|z_{2}\right|=3, z_{1} \overline{z_{2}}+\overline{z_{1}} z_{2}=-9$, and
$$
\left|z_{1} \overline{z_{2}}\right|+\left|\overline{z_{1}} z_{2}\right|=9
$$

Let $z_{1} \overline{z_{2}}=9(\cos \theta+i \sin \theta)$,
then $\overline{z_{1}} z_{2}=9(\cos \theta-i \sin \theta)$.
From $-9=z_{1} \overline{z_{2}}+\overline{z_{1}} z_{2}=18 \cos \theta$, we get
$$
\cos \theta=-\frac{1}{2} \text {. }
$$

Thus, $z_{1} \overline{z_{2}}=9 \omega$ or $z_{1} \overline{z_{2}}=9 \omega^{2}$, where
$$
\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} i .
$$

When $z_{1} \overline{z_{2}}=9 \omega$, we have

Therefore,
$$
\begin{array}{c}
\left(z_{1} \overline{z_{2}}\right)^{2000}+\left(\overline{z_{1}} z_{2}\right)^{2000}=-9^{2000}, \\
\log _{3}\left|\left(z_{1} \overline{z_{2}}\right)^{2000}+\left(\overline{z_{1}} z_{2}\right)^{2000}\right|=4000 .
\end{array}
$$

When $\overline{z_{1}} z_{2}=9 \omega^{2}$, we get the same result, so the answer is 4000.