If $F : Ab \rightarrow Ab$ is an additive functor and $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ is an exact split sequence then ... 
If $F : Ab \rightarrow Ab$ is an additive functor and $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ is an exact split sequence $\Rightarrow$ $0 \rightarrow FA \rightarrow FB \rightarrow FC \rightarrow 0$ is an exact split sequence.

Split:  Since $s: C \rightarrow B$ exists such that $ps = 1_C$ ($p: B \rightarrow C)$, it follows that $Fs : FC \rightarrow FB$ exists and $Fp \circ Fs = 1_{FC}$ by the definition of a functor.
Exact: I know I have to show that $\text{im $Fi$} = \ker Fp$ $(i : A \rightarrow B)$, but I'm having trouble figuring out a way to do this.

Question (1) : Is the proof I gave of the second sequence being split correct since I didn't use the fact that $F$ is an additive functor at all?
Question (2): Anyone have any ideas as to how to show this?
 A: For a sequence $0\to A\to B\to C\to0$ to be split exact, it is necessary
and sufficient that morphism $B\to A$ and $C\to B$ exist, satisfying
a few identities together with the given maps $A\to B$ and $B\to C$.
Any additive functor will preserve these identities, so map the
split exact sequence to another one.
A: For the first question, we need the additivity. In fact, apply the functor $F$ to the sequence, then you have
$F0 \rightarrow FA \rightarrow FB \rightarrow FC \rightarrow F0$. So the question is, why do we have $0$ and not $F0$? This comes from the additivity of $F$ because $F$ sends the zero function on the zero function. So there is no problem to write $0$.
PS : my first answer here, excuse my English.
A: I think it is nice to use splitting lemma here. It tells you that a split short exact sequence $$ A\to B \to C$$ is equivalent to
$$ A\to A\oplus C \to C  $$
where the maps are the inclusion are projection.
It is easy to see that such sequences sequences are always split short exact.
So what you need it hat additive functors preserve splittings and direct sums with its inclusion/projections. The first thing is easy since any functor preserves splittings.(As you said correctly) For the second part it depends on your definition of additive. (Preserving biproducts is equivalent to be additive)
