Endomorphism of Abelian variety over finite field

Question

Let $A$ be an abelian variety defined over a finite field $k$ of characteristic $p$, such that $A/k$ is simple but not absolutely simple.

Let $f\in End_k(A)$ be an endomorphism defined over $k$, and $g\in End(A)$ is an endomorphism defined over an algebraic closure of $k$.

1. We have that either $f=0$, either $\ker f$ is finite (equivalently $f$ is surjective)?

From now, suppose that $f\neq 0$

We have that $g\circ f=0$ iff $g=0$

2. If $f\circ g=0$, which conditions on $f$ are necessary to have $g=0$?

I'am looking for a necessary condition. Sufficient conditions to have $g=0$ are for example:

(i) $f$ commutes with $g$, (ii) $g$ is defined over $k$,..

3. Is there a relation between the question 2. and the fact that $p$ divides or not the cardinality fo $\ker f$?

Answer 1

(1) is true, because if f is not $0$ then $(\ker f)^0$ defines a proper subabelian variety over $k$, which is trivial if $A$ is simple.

(2) If $fg = 0$ the $g$ factors through the finite subscheme $\ker f$. On the other hand, its image is connected and reduced, so is already trivial. In summary, $g=0$ is sufficient and necessary.