Functions

The denominator must not be zero, and $x + 4 = 0$ when $x = -4$.

$x^2$ is never negative, so its smallest value is 0, giving a smallest $y$ of 3.

Apply $f$ first: $f(x) = x+1$. Then $g(x+1) = 3(x+1) = 3x+3$. Note $fg(x) = 3x + 1$, which is different.

From $y = \dfrac{x}{4} - 2$, add 2 and multiply by 4: $x = 4(y+2)$. Check: $f(8) = 0$ and $f^{-1}(0) = 8$ ✓

The expression under the root must be non-negative: $5 - x \ge 0$, so $x \le 5$. Note the inequality reverses when $x$ is moved across.

$\sin 0 = \sin \pi = 0$, so different inputs share an output and the horizontal line test fails. Restricting to $-\tfrac{\pi}{2} \le x \le \tfrac{\pi}{2}$ makes the inverse sine possible.