(1) Since the terminal side of angle $\theta$ passes through point $P(-4,3)$, we can find $\sin \theta$, $\cos \theta$, and $\tan \theta$ using the coordinates of point $P$.

The distance $r$ between point $P$ and the origin $O(0,0)$ is given by:

$$r = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$

Now, we can find the sine, cosine, and tangent of angle $\theta$:

$$\sin \theta = \frac{y}{r} = \frac{3}{5}$$

$$\cos \theta = \frac{x}{r} = -\frac{4}{5}$$

$$\tan \theta = \frac{y}{x} = -\frac{3}{4}$$

(2) Now let's find the given expression:

$$\frac{\cos (\theta - \frac{\pi}{2})}{\sin (\frac{\pi}{2} + \theta)} \sin (\theta + \pi) \cos (2\pi - \theta)$$

First, we can simplify the trigonometric expressions inside the parentheses using the cofunction identities:

$$\cos \left(\theta - \frac{\pi}{2}\right) = \sin \theta$$

$$\sin \left(\frac{\pi}{2} + \theta\right) = \cos \theta$$

Now we can rewrite the expression as:

$$\frac{\sin \theta}{\cos \theta} \sin (\theta + \pi) \cos (2\pi - \theta)$$

Next, we can use the identity $\sin(\theta + \pi) = -\sin\theta$ and the fact that $\cos(2\pi - \theta) = \cos\theta$:

$$\frac{\sin \theta}{\cos \theta} (-\sin \theta) \cos \theta$$

Finally, we can cancel out the common factors and get:

$$- \sin^2 \theta$$

Substitute the value of $\sin \theta = \frac{3}{5}$:

$$- \left(\frac{3}{5}\right)^2 = -\frac{9}{25}$$

So the final answer is: $\boxed{-\frac{9}{25}}$.