Let's go through the problems one by one.

1. Prove that $\sin^{-1} \left(\frac{4}{5}\right) + \cos^{-1} \left(\frac{3}{5}\right) = \frac{\pi}{2}$.

Solution:

Given the equation: $\sin^{-1} \left(\frac{4}{5}\right) + \cos^{-1} \left(\frac{3}{5}\right)$

We know that: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$

So, if $\sin^{-1} \left(\frac{4}{5}\right) = \theta$, then $\sin(\theta) = \frac{4}{5}$. Therefore, by the Pythagorean identity $\cos^2 \theta = 1 - \sin^2 \theta$, we have: $\cos(\theta) = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5}$

This means: $\cos^{-1} \left(\frac{3}{5}\right) = \frac{\pi}{2} - \sin^{-1} \left(\frac{4}{5}\right)$ Thus: $\sin^{-1} \left(\frac{4}{5}\right) + \cos^{-1} \left(\frac{3}{5}\right) = \frac{\pi}{2}$ Proven.

2. Find the angle between $5i$ and $2 + 2i$.

Solution:

The angle of a complex number $z = x + yi$ with the positive real axis is given by $\theta = \arg(z) = \tan^{-1} \left(\frac{y}{x}\right)$.

For $5i$: $\theta_1 = \tan^{-1} \left(\frac{5}{0}\right) = \frac{\pi}{2}$

For $2 + 2i$: $\theta_2 = \tan^{-1} \left(\frac{2}{2}\right) = \tan^{-1}(1) = \frac{\pi}{4}$

The angle between them is: $\theta = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$

3. Solve for $x$ in $15 - 6\left(\frac{12 - x}{3}\right) = 4x$.

Solution:

Start by simplifying the equation: $15 - 6\left(\frac{12 - x}{3}\right) = 4x$ $15 - 6\left(4 - \frac{x}{3}\right) = 4x$ $15 - (24 - 2x) = 4x$ $15 - 24 + 2x = 4x$ $-9 + 2x = 4x$ $-9 = 2x$ $x = -\frac{9}{2}$

4. For what values of $x$ is the matrix (A = \begin{pmatrix} 1 & 2