18. (16 points) 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 $|B M| \cdot|B N|$.
18. (1) Establish a Cartesian coordinate system with the line $AB$ as the $x$-axis and the perpendicular bisector of $AB$ as the $y$-axis. Let $|CA| + |CB| = 2a (a > 3)$ be a constant, then the locus of point $C$ is an ellipse with foci at $A$ and $B$. Therefore, the focal distance is $2c = |AB| = 6$.
Since $\cos C = \frac{|CB|^2 + |CA|^2 - 6^2}{2|CB| \cdot |CA|}$
$$
\begin{array}{l}
= \frac{(|CB| + |CA|)^2 - 2|CB| \cdot |CA| - 36}{2|CB| \cdot |CA|} \\
= \frac{2a^2 - 18}{|CB| \cdot |CA|} - 1,
\end{array}
$$

Also, $|CB| \cdot |CA| \leq \left(\frac{2a}{2}\right)^2 = a^2$, so $\cos C \geq 1 - \frac{18}{a^2}$.
From the problem, we have $1 - \frac{18}{a^2} = \frac{7}{25}$, which means $a^2 = 25$.
At this point, $|PA| = |PB|$, and the coordinates of point $P$ are $(0, \pm 4)$. Therefore, the equation of the locus of point $C$ is
$$
\frac{x^2}{25} + \frac{y^2}{16} = 1 (y \neq 0).
$$
(2) Suppose $A(-3,0)$, $M(x_1, y_1)$, and $N(x_2, y_2)$.

When the inclination angle of line $MN$ is not 900, let its equation be $y = k(x + 3)$, and substitute it into the ellipse equation to simplify, 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
$$
\begin{array}{l}
|\boldsymbol{B}\boldsymbol{M}| \cdot |\boldsymbol{B}\boldsymbol{N}| = \left(5 - \frac{3}{5} x_1\right)\left(5 - \frac{3}{5} x_2\right) \\
= 25 + \frac{450k^2}{16 + 25k^2} + \frac{81k^2 - 144}{16 + 25k^2} \\
= 25 + \frac{531k^2 - 144}{16 + 25k^2} = 25 + \frac{531}{25} \cdot \frac{k^2 - \frac{144}{531}}{k^2 + \frac{16}{25}}.
\end{array}
$$

Thus, we only need to consider the minimum value of $\frac{k^2 - \frac{144}{531}}{k^2 + \frac{16}{25}} = 1 - \frac{\frac{16}{25} + \frac{144}{531}}{k^2 + \frac{16}{25}}$.
Clearly, when $k = 0$, $|\boldsymbol{B}\boldsymbol{M}| \cdot |\boldsymbol{B}\boldsymbol{N}|$ takes the minimum value 16.
When the inclination angle of line $MN$ is $90^\circ$, $x_1 = x_2 = -3$, we get $|BM| \cdot |BN| = \left(\frac{34}{5}\right)^2 > 16$.
But $\frac{x^2}{25} + \frac{y^2}{16} = 1 (y \neq 0)$, so $k \neq 0$.
Such $M$ and $N$ do not exist, i.e., the set of minimum values of $|BM| \cdot |BN|$ is empty.