13.4. 19 ** In $\triangle A B C$ with a fixed perimeter, it is known that $|A B|=6$, and when vertex $C$ is at a fixed point $P$, $\cos C$ has a minimum value of $\frac{7}{25}$.
（1）Establish an appropriate coordinate system and find the equation of the locus of vertex $C$;
(2) Draw a line through point $A$ that intersects the curve from (1) at points $M$ and $N$, and find the minimum value of $|\overrightarrow{B M}| \cdot|\overrightarrow{B N}|$.
(1) Taking the line $AB$ as the $x$-axis and the perpendicular bisector of segment $AB$ as the $y$-axis, establish a Cartesian coordinate system. Let $|CA| + |CB| = 2a (a > 3)$ be a constant, so the locus of point $C$ is an ellipse with $A$ and $B$ as foci. Therefore, the focal distance $2c = |AB| = 6$. Since
$$
\begin{aligned}
\cos C = & \frac{|CA|^2 + |CB|^2 - 6^2}{2|CA||CB|} = \frac{(|CA| + |CB|)^2 - 2|CA||CB| - 36}{2|CA||CB|} \\
& = \frac{2a^2 - 18}{|CA||CB|} - 1,
\end{aligned}
$$

and $|CA| \cdot |CB| \leq \left(\frac{2a}{2}\right)^2 = a^2$, so $\cos C \geq 1 - \frac{18}{a^2}$. Given that $1 - \frac{18}{a^2} = \frac{7}{25}$, we have $a^2 = 25$. At this point, $|PA| = |PB|$, and the coordinates of point $P$ are $P(0, \pm 4)$. Therefore, the equation of the locus of point $C$ is $\frac{x^2}{25} + \frac{y^2}{16} = 1 (y \neq 0)$.
(2) Without loss of generality, let the coordinates of point $A$ be $A(-3, 0)$, and the coordinates of points $M$ and $N$ be $M(x_1, y_1)$ and $N(x_2, y_2)$, respectively. When the inclination angle of line $MN$ is not $90^\circ$, let its equation be $y = k(x + 3)$. Substituting into the ellipse equation and simplifying, we get
$$
\left(\frac{1}{25} + \frac{k^2}{16}\right) x^2 + \frac{3}{8} k^2 x + \left(\frac{9k^2}{16} - 1\right) = 0,
$$

Clearly, $\Delta \geq 0$, so
$$
x_1 + x_2 = -\frac{150k^2}{16 + 25k^2}, \quad x_1 x_2 = \frac{225k^2 - 400}{16 + 25k^2},
$$

By the second definition of the ellipse, we have
$$
\begin{aligned}
|\overrightarrow{BM}| \cdot |\overrightarrow{BN}| & = \left(5 - \frac{3}{5} x_1\right)\left(5 - \frac{3}{5} x_2\right) = 25 - 3(x_1 + x_2) + \frac{9}{25} x_1 x_2 \\
& = 25 + \frac{450k^2}{16 + 25k^2} + \frac{81k^2 - 144}{16 + 25k^2} = 25 + \frac{531k^2 - 144}{16 + 25k^2} \\
& = 25 + \frac{531}{25} \cdot \frac{k^2 - \frac{144}{531}}{k^2 + \frac{16}{25}}
\end{aligned}
$$

We need to consider the minimum value of $\frac{k^2 - \frac{144}{531}}{k^2 + \frac{16}{25}}$, which is equivalent to considering the minimum value of $1 - \frac{\frac{16}{25} + \frac{144}{531}}{k^2 + \frac{16}{25}}$. Clearly, when $k = 0$,
$|\overrightarrow{BM}| \cdot |\overrightarrow{BN}|$ takes the minimum value 16.
When the inclination angle of line $MN$ is $90^\circ$, $x_1 = x_2 = -3$, and we get $|\overrightarrow{BM}| \cdot |\overrightarrow{BN}| = \left(\frac{34}{5}\right)^2 > 16$. However, $\frac{x^2}{25} + \frac{y^2}{16} = 1 (y \neq 0)$, so $k \neq 0$, and such points $M$ and $N$ do not exist. Therefore, the set of minimum values of $|\overrightarrow{BM}| \cdot |\overrightarrow{BN}|$ is empty.