5. In a regular tetrahedron $P-ABC$, points $E$ and $F$ are on edges $PB$ and $PC$, respectively. If $PE \neq PF$, and $AE = AF = 2$, $EF = \sqrt{3}$, then the volume of the tetrahedron $P-AEF$ is $\qquad$
5. $\frac{1}{8}$.

As shown in Figure 4.
By the problem, let's assume the height of the regular tetrahedron is $h$,
$$
\begin{array}{l}
P F=B E=x, \\
P E=F C=y .
\end{array}
$$

Then the edge length is $x+y$.
Thus, in $\triangle P E F$,
$$
x^{2}+y^{2}-x y=3 \text {; }
$$

In $\triangle P A F$, $x^{2}+y^{2}+x y=4$.
From the above two equations, we get $x+y=\frac{3 \sqrt{2}}{2}, x y=\frac{1}{2}$.
Then $S_{\triangle P E F}=\frac{1}{2} x y \sin \frac{\pi}{3}=\frac{\sqrt{3}}{8}$,
$$
h=(x+y) \frac{\sqrt{6}}{3}=\sqrt{3} \text {. }
$$

Therefore, $V_{\text {tetrahedron } A-P E F}=\frac{1}{3} \times \frac{\sqrt{3}}{8} \times \sqrt{3}=\frac{1}{8}$.