Problem 30

1. Since $\overrightarrow{a}$ is parallel to $\overrightarrow{b}$, we have:

$$\frac{3}{2} = \frac{-4}{x} \Rightarrow x = -\frac{8}{3}.$$

And since $\overrightarrow{a}$ is perpendicular to $\overrightarrow{c}$, we have:

$$3 \cdot 2 + (-4) \cdot y = 0 \Rightarrow y = \frac{3}{2}.$$

Thus, $\overrightarrow{b} = (2, -\frac{8}{3})$ and $\overrightarrow{c} = (2, \frac{3}{2})$.

Now, we can compute the dot product of $\overrightarrow{b}$ and $\overrightarrow{c}$:

$$\overrightarrow{b} \cdot \overrightarrow{c} = 2 \cdot 2 + \left( -\frac{8}{3} \right) \cdot \frac{3}{2} = \boxed{0}.$$

2. Let $\theta$ be the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$. We can find $\cos{\theta}$ using the dot product formula:

$$\cos{\theta} = \frac{\overrightarrow{b} \cdot \overrightarrow{c}}{|\overrightarrow{b}||\overrightarrow{c}|} = \frac{0}{\sqrt{2^2 + \left(-\frac{8}{3}\right)^2}\sqrt{2^2 + \left(\frac{3}{2}\right)^2}} = 0.$$

Since $0^\circ \leqslant \theta \leqslant 180^\circ$, we have:

$$\theta = \boxed{90^\circ}.$$