5. Given a regular tetrahedron $P-ABC$ with edge length $1$, points $E, F, G, H$ are on edges $PA, PB, CA, CB$ respectively. If $\overrightarrow{EF}+\overrightarrow{GH}=\overrightarrow{AB}$, and $\overrightarrow{EH} \cdot \overrightarrow{FG}=\frac{1}{18}$, then $\overrightarrow{EG} \cdot \overrightarrow{FH}=$ $\qquad$ .
Given by Wang Zheng: $\frac{5}{18}$.
From $\overrightarrow{E F}+\overrightarrow{G H}=\overrightarrow{A B}$, we know that the vectors $\overrightarrow{E F}$, $\overrightarrow{G H}$, and $\overrightarrow{A B}$ are coplanar. If $\overrightarrow{G H}$ is not parallel to $\overrightarrow{A B}$, then both $\overrightarrow{G H}$ and $\overrightarrow{A B}$ lie in the plane $ABC$, which implies that $\overrightarrow{E F}$ is parallel to the plane $ABC$, leading to $\overrightarrow{E F} \| \overrightarrow{A B}$, and thus $\overrightarrow{G H} \| \overrightarrow{A B}$, which is a contradiction. Therefore, $\overrightarrow{G H} \| \overrightarrow{A B}$. Similarly, we can conclude that $\overrightarrow{E F} \| \overrightarrow{A B}$.

Thus, we can set $|A E|=k$, then $|A G|=1-k$, which implies that quadrilateral $E F H G$ is an isosceles trapezoid with $|E F|=1-k$, $|G H|=k$, and $|E G|=|F H|=\sqrt{3 k^{2}-3 k+1}$. By the Pythagorean theorem, the height of the trapezoid is $h=\sqrt{2 k^{2}-2 k+\frac{3}{4}}$. Therefore, $\overrightarrow{E H} \cdot \overrightarrow{F G}=h^{2}-\left(\frac{|E F|+|G H|}{2}\right)^{2}=h^{2}-\frac{1}{4}=\frac{1}{18}$, so $h^{2}=\frac{11}{36}$. Hence,
$$
\overrightarrow{E G} \cdot \overrightarrow{F H}=h^{2}-\left(\frac{|E F|-|G H|}{2}\right)^{2}=h^{2}-\left(k^{2}-k+\frac{1}{4}\right)=\frac{h^{2}}{2}+\frac{1}{8}=\frac{5}{18} .
$$