In ZF, if injection from A to B doesn't exist, then does surjection from A to B exist?

In fact, this statement is equivalent to the axiom of choice. Given a nonempty set $B$, let $A$ be its Hartogs number, the least ordinal that does not inject into $B$. Then, by your assumption, there must be a surjection $f:A\to B$. But since $A$ is well-ordered, we can use this to construct an injection $g:B\to A$, by letting $g(b)$ be the least $a$ such that $f(a)=b$. This gives a well-ordering of $B$.

Let $A$ be an amorphous set; then there is no injection from $\Bbb N$ to $A$, and there is no surjection from $\Bbb N$ to $A$.

Let $f:\Bbb N\to A$. If $f$ were injective, then $A_0=\{f(2n):n\in\Bbb N\}$ and $A\setminus A_0$ would be disjoint infinite subsets of $A$ whose union is $A$, which is impossible, since $A$ is amorphous. And if $f$ were surjective, the function

$$h:A\to\Bbb N:a\mapsto\min\{n\in\Bbb N:f(n)=a\}$$

would be a bijection from $A$ to the infinite set $M=h[A]\subseteq\Bbb N$. If $\varphi:\Bbb N\to M$ is the canonical order-isomorphism, $h^{-1}\circ\varphi$ would be a bijection from $\Bbb N$ to $A$, which again is impossible.