Preimage of a compact set

Not generally. The pre-image is closed because $f$ is continuous. But it need not be bounded. The simplest example is to let $f(x,y)=0$ for all $x, y.$ Then $\{0\}$ is compact and $f^{-1}\{0\}=\mathbb R^2.$

It's not true. not even for $\mathbb R $ to $\mathbb R$. Consider $\sin(x)$

$f(x_1, ... ,x_n) = \sin(5x_1)$ is an example of a continuous $f$ with unbounded preimage of $f([0,1])$.