18. 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 $|\overrightarrow{B M}| \cdot|\overrightarrow{B N}|$.
18. (1) Taking the line $AB$ as the $x$-axis and the perpendicular bisector of segment $AB$ as the $y$-axis, we establish a Cartesian coordinate system. Let $|CA| + |CB| = 2a (a > 3)$ be a constant, so the locus of point $C$ is an ellipse with foci at $A$ and $B$. Therefore, the foci distance $2c = AB = 6$.
Since $\cos C = \frac{|CB|^2 + |CA|^2 - 6^2}{2|CB| \cdot |CA|} = \frac{(|CB| + |CA|)^2 - 2|CB| \cdot |CA| - 36}{2|CB| \cdot |CA|} = \frac{2a^2 - 18}{|CB| \cdot |CA|} - 1$,
and $|CB| \cdot |CA| \leq \left(\frac{2a}{2}\right)^2 = a^2$, we have $\cos C \geq 1 - \frac{18}{a^2}$. Given that $1 - \frac{18}{a^2} = \frac{7}{25}$, we find $a^2 = 25$. 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) Suppose the coordinates of point $A$ are $A(-3, 0)$, and the coordinates of points $M$ and $N$ are $M(x_1, y_1)$ 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 $x_1 + x_2 = -\frac{150k^2}{16 + 25k^2}$ and $x_1 x_2 = \frac{225k^2 - 400}{16 + 25k^2}$.
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 $k = 0$, $|\overrightarrow{BM}| \cdot |\overrightarrow{BN}|$ achieves its minimum value of 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.