Presheaf image of a monomorphism of sheaves is a sheaf EDIT: The original version regarding coker is false.
Consider a morphism of sheaves of abelian groups $\Phi:\mathscr{F}\to\mathscr{G}$. 
It is not in general true that Im$_{pre}\Phi$ is a sheaf. However, if $\Phi$ is a monomorphism, then $\Phi_U$ is a monomorphism on open sets and Im$_{pre}\Phi$ seems to be a sheaf. Consider the explicit construction of Im$_{pre}\Phi$ where
$$\text{Im}_{pre} \Phi(U) := \text{Im}\; \Phi_U \subset \mathscr{G}(U)$$
 since we are dealing with sheaves of abelian groups, this seems to be a sheaf.  
Is this true? What if we are dealing with a monomorphism $\Phi:\mathscr{F}\to\mathscr{G}$ of sheaves of some unknown abelian category, is it still true that Im$_{pre} \Phi$ is a sheaf? And if so what is a purely abstract argument?
 A: Perhaps I've misunderstood what you wrote, but it seems to me that your statement about the presheaf cokernel being a sheaf is false.
E.g. take $X := \mathbb C^{\times}$ with its usual topology, let $\mathscr G := \mathscr O_X$ denote the sheaf of holomorphic functions on $X$, and let $\mathscr F$ denote the sheaf of locally constant functions on $X$ which take values in $2\pi i \mathbb Z.$  
Then $\mathscr F \hookrightarrow \mathscr G$ (any locally constant function is holomorphic).  
Let $\mathscr O_X^{\times}$ denote the sheaf of nowhere-zero holomorphic functions on $X$ (which is a sheaf of abelian groups via multiplication).  
There is a natural isomorphism $\mathscr G/\mathscr F \buildrel \sim \over\longrightarrow \mathscr O_X^{\times}$,
given via the morphism $f \mapsto e^f.$
This map is not surjective on global sections: there is no global section
of $\mathscr G$ which maps to the global section $z$ of $\mathscr O_X^{\times}.$
(Concretely, there is no single valued branch of $\log z$ defined on $X$.)
Thus $\mathscr G(U) \to (\mathscr G/\mathscr F)(U)$ is not surjective, and in particular, the natural injection
$$\mathscr G(U)/\mathscr F(U) \to (\mathscr G/\mathscr F)(U)$$
is not surjective. 
(This failure of surjectivity is the source of sheaf cohomology.  In the particular case I've described here, the related cohomological fact is that the $1$st cohomology of $X$ with integral coefficients is non-zero.)
