5. Given that $A$ and $B$ are the left and right vertices of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, $F_{1}$ and $F_{2}$ are the left and right foci, respectively, and $C$ is a point on the hyperbola different from $A$ and $B$. If the lines $A C$ and $B C$ intersect the right directrix of the hyperbola at points $M$ and $N$, respectively, then $\overrightarrow{F_{1} M} \cdot \overrightarrow{F_{2} N}$ $=$ . $\qquad$
5. $-2 b^{2}$.

By symmetry, without loss of generality, assume point $C\left(x_{C}, y_{C}\right)$ is in the first quadrant.
Thus, $M\left(\frac{a^{2}}{c}, y_{M}\right), N\left(\frac{a^{2}}{c}, y_{N}\right)$.
From the collinearity of three points, we get
$$
\begin{array}{l}
y_{M}=\frac{\frac{a^{2}}{c}+a}{x_{C}+a} y_{C}, y_{N}=-\frac{a-\frac{a^{2}}{c}}{x_{C}-a} y_{C} \\
\Rightarrow y_{M} y_{N}=\frac{\left(\frac{a^{4}}{c^{2}}-a^{2}\right) y_{C}^{2}}{x_{C}^{2}-a^{2}}=-\frac{b^{4}}{c^{2}} . \\
\text { Therefore, } \overrightarrow{F_{1} M} \cdot \overrightarrow{F_{2} N}=\left(\frac{a^{2}}{c}+c, y_{M}\right) \cdot\left(\frac{a^{2}}{c}-c, y_{N}\right) \\
=\frac{a^{4}}{c^{2}}-c^{2}+y_{M} y_{N}=-2 b^{2} .
\end{array}
$$