Q1: Identify the function from its graph, $\sin^{-1}(2x)$

The domain is the set of $x$-values the curve covers. It starts on the dashed line $x = -\frac{1}{2}$ and stops on the dashed line $x = \frac{1}{2}$⁠.

The range is the set of $y$-values. The curve rises from $-\frac{\pi}{2}$ at its left end to $\frac{\pi}{2}$ at its right end.

$\sin^{-1}u$ is defined only when $-1 \le u \le 1$⁠. Here the input is $u = 2x$⁠.

Dividing all three parts by 2, a positive number, keeps the signs $\le$ the same way.

As $2x$ runs over all of $[-1, 1]$⁠, $\sin^{-1}(2x)$ runs over all of $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠. So the range matches too.

For $\sin^{-1}\left(\frac{x}{2}\right)$ the input is $\frac{x}{2}$⁠:

Watch out: $2\sin^{-1}x$ multiplies the output by 2. Its domain stays $[-1, 1]$ and its range becomes $[-\pi, \pi]$⁠.

$\cos^{-1}u$ is defined only when $-1 \le u \le 1$⁠. Put $u = 3x$ and divide all three parts by 3.

Changing the input from $x$ to $3x$ shrinks the domain by the factor 3. Changing it to $\frac{x}{3}$ stretches it by 3:

Watch out: $3\cos^{-1}x$ multiplies the output by 3, which changes the range, not the domain.