The problem states that $5\cos{A} + 3 = 0$ and $180^\circ < A < 360^\circ$. We need to determine the values of $\tan^2{A}$ and $\frac{\sin{A}}{\cos{A}}$.

1. Problem Description

The problem states that $5\cos{A} + 3 = 0$ and $180^\circ < A < 360^\circ$. We need to determine the values of $\tan^2{A}$ and $\frac{\sin{A}}{\cos{A}}$.

2. Solution Steps

First, we solve for $\cos{A}$:

$5\cos{A} + 3 = 0$

$5\cos{A} = -3$

$\cos{A} = -\frac{3}{5}$

Since $180^\circ < A < 360^\circ$, $A$ is in the third or fourth quadrant. Since $\cos{A} = -\frac{3}{5} < 0$, $A$ must be in the third quadrant. Therefore, $\sin{A} < 0$ in this quadrant.

We use the Pythagorean identity $\sin^2{A} + \cos^2{A} = 1$ to find $\sin{A}$:

$\sin^2{A} + \left(-\frac{3}{5}\right)^2 = 1$

$\sin^2{A} + \frac{9}{25} = 1$

$\sin^2{A} = 1 - \frac{9}{25}$

$\sin^2{A} = \frac{16}{25}$

$\sin{A} = \pm\frac{4}{5}$

Since $A$ is in the third quadrant, $\sin{A} < 0$, so $\sin{A} = -\frac{4}{5}$.

Now we can find $\tan{A}$:

$\tan{A} = \frac{\sin{A}}{\cos{A}} = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3}$

(a) $\tan^2{A} = \left(\frac{4}{3}\right)^2 = \frac{16}{9}$

(b) $\frac{\sin{A}}{\cos{A}} = \tan{A} = \frac{4}{3}$

3. Final Answer

(a) $\tan^2{A} = \frac{16}{9}$

(b) $\frac{\sin{A}}{\cos{A}} = \frac{4}{3}$