1) $f\colon \mathbb{Q} \to \mathbb{Q}$, $f(x) = 1 + 2x$.
For $f$ to be surjective it must be the case that every $y \in \mathbb Q$ can be written as $y = f(x)$ for some $x$, but $x = \frac{y - 1}{2}$ does. So $y$ is surjective.
2) $f\colon \mathbb{Z} \to \mathbb{N}\cup\{0\}$, $f(x) = |1 - x|$.
let $y \in \mathbb{N}\cup\{0\}$ then $1-y \in \mathbb Z$ and $y = f(1-y)$ so $f$ is surjective.
3) $f\colon \mathbb{Q} \to \mathbb{Q}$, $f(x) = 4 - 2x^3$.
This is the most interesting one, because of the cube I don't expect it to be surjective. Note that $2-f(x)/2 = x^3$ so if $f(x)$ really is surjective then take any rational $q$ and $2-q/2$ is a cube! Obviously this is false so $f$ is proved not surjective. For a concrete example take any $q$ such that $2-q/2$ is not a cube say $q = 2$ then there's no $x$ such that $q=f(x)$.
4) $f\colon \mathbb{Z}^2 \to \mathbb{Z}$, $f(x, y) = x + y$.
This is easily surjective because $f(x,0)$ is the identity function, which is surjective.