5. The distance from a vertex of the base of a regular tetrahedron to the centroid of the opposite face is 4. The maximum volume of this regular tetrahedron is $\qquad$
5.18.

As shown in Figure 3, in the regular tetrahedron $P-ABC$, let the centroid of the lateral face $PBC$ be $G$, and extend $PG$ to intersect $BC$ at $D$. Clearly, $D$ is the midpoint of $BC$. Let the projections of $P$ and $G$ on the base be $O$ and $H$, respectively, with the base edge length being $a$ and the height of the tetrahedron being $h$.
$$
\because AG=4 \text {, }
$$

and $GH=\frac{1}{3} PO=\frac{h}{3}$,
$$
HD=\frac{1}{3} OD=\frac{1}{9} AD=\frac{\sqrt{3}}{18} a \text {, }
$$

then $AH=\frac{4 \sqrt{3}}{9} a$.
$$
\therefore\left(\frac{4 \sqrt{3}}{9} a\right)^{2}+\left(\frac{h}{3}\right)^{2}=4^{2} \text { , }
$$

which gives
$$
\begin{array}{l}
16 a^{2}+3 h^{2}=16 \times 27, \\
8 a^{2}+8 a^{2}+3 h^{2}=16 \times 27 . \\
16 \times 27 \geqslant 3 \sqrt[3]{8 a^{2} 8 a^{2} 3 h^{2}}, a^{2} h \leqslant \frac{216}{\sqrt{3}},
\end{array}
$$

Thus, $V_{P-ABC}=\frac{1}{3} \cdot \frac{\sqrt{3}}{4} a^{2} h \leqslant \frac{\sqrt{3}}{12} \times \frac{216}{\sqrt{3}}=18$.