Solution Detailed Steps:

(1) Given $\cos B - \cos \frac{A+C}{2} = 0$,

Since $A + B + C = \pi$ in a triangle, we can rewrite $A + C$ as $\pi - B$. Hence,
$$\cos B - \cos \left(\frac{\pi - B}{2}\right) = 0$$

This simplifies to:
$$\cos B - \sin \left(\frac{B}{2}\right) = 0$$

Recall the double angle identity for sine, $\sin^2 x = \frac{1 - \cos(2x)}{2}$. Applying it with $x = \frac{B}{2}$ gives:
$$\cos B - 1 + 2\sin^2\left(\frac{B}{2}\right) = 0$$

Rearranging and solving for $\sin\left(\frac{B}{2}\right)$:
$$2\sin^2\left(\frac{B}{2}\right) + \sin\left(\frac{B}{2}\right) - 1 = 0$$

$$\sin \left(\frac{B}{2}\right) = \frac{-1 + \sqrt{1 + 8}}{4} = \frac{1}{2}$$ or $\sin \left(\frac{B}{2}\right) = -1$ (which is discarded as $\sin$ can't be $-1$ for $0 0$, then we have $a = 3m$.

From the formula for the area of a triangle, $\frac{1}{2}ab\sin C = \text{Area}$, and given the altitude from $AC$ is $\frac{12\sqrt{3}}{7}$, we have:
$$\frac{1}{2}ac\sin B = \frac{1}{2} \times b \times \frac{12\sqrt{3}}{7}$$
Since $B = \frac{\pi}{3}$, $\sin B = \frac{\sqrt{3}}{2}$, this becomes:
$$\frac{1}{2} \times 3m \times 8m \times \frac{\sqrt{3}}{2} = \frac{1}{2} \times b \times \frac{12\sqrt{3}}{7}$$
Simplifying, we find $b = 7m^2$.

Using the cosine rule $b^2 = a^2 + c^2 - 2ac\cos B$, and knowing $B = \frac{\pi}{3}$, $\cos B = \frac{1}{2}$, we have:
$$49m^4 = 9m^2 + 64m^2 - 2 \times 3m \times 8m \times \frac{1}{2}$$
$$49m^4 = 73m^2 - 24m^2$$
$$49m^4 = 49m^2$$
Solving for $m$, we find $m = 1$.

Therefore, $a = 3$, $b = 7$, and $c = 8$.

Hence, the perimeter of $\triangle ABC$ is $a + b + c = 3 + 7 + 8 = \boxed{18}$.