5. For the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1(0<b<4)$, the left focus is $F$, and line $l$ intersects the ellipse at points $A$ and $B$, with $|A B|$ $=6$. Then the maximum value of $|F A| \cdot|F B|$ is $\qquad$
5. 25 .

Let the right focus of the ellipse be $F^{\prime}$, and connect $A F^{\prime}$ and $B F^{\prime}$. Then,
$$
|A B| \leqslant\left|A F^{\prime}\right|+\left|B F^{\prime}\right| \text {. }
$$

When the chord $A B$ passes through the right focus $F^{\prime}$, the equality holds.
Thus, $|F A|+|F B|+|A B|$
$$
\begin{array}{l}
\leqslant|F A|+\left|F^{\prime} A\right|+|F B|+\left|F^{\prime} B\right| \\
=2 a+2 a=4 a=16 .
\end{array}
$$

Therefore, $|F A|+|F B| \leqslant 10$.
Hence, $|F A| \cdot|F B| \leqslant\left(\frac{|F A|+|F B|}{2}\right)^{2} \leqslant 25$.
For the equalities in (1) and (2) to hold simultaneously, $A B$ should pass through the right focus and be perpendicular to the $x$-axis.
Thus, $(|F A| \cdot|F B|)_{\max }=25$.