Since $G$ has odd order, that means $G$ is finite. So $\phi$ can be injective if and only if it is surjective. To formally prove surjectivity, argue by contradiction. Suppose there is a $y \in G$ such that $\forall{x} \in G$, $\phi(x) \neq y$. That implies there are two distinct $x_{1}, x_{2}$ such that $\phi(x_{1}) = \phi(x_{2})$ by injectivity.
Note: I'm using additive notation for both groups G and H, however, I assume ONLY $H$ abelian, I will not use abelian for $G$, tell me if I mistake that. but it totally can be done without using $G$ abelian.
Proof: Let $A=Hom(G,H)$, with the sum $(f+g)(x)=f(x)+g(x)$. Let's prove $A$ it's a abelian group:
Closed: $(f+g)(x+y)=f(x+y)+g(x+y)=f(x)+f(y)+g(x)+g(y)$
Now IMPORTANT, we use "$H$ is abelian" hypothesis, if not, we can't do this step, it's simply false.
$=f(x)+g(x)+f(y)+g(y)=(f+g)(x)+(f+g)(y)$ that proves the sum of homomorphism is homomorphism, so it's closed.
Associative: I copy and improve your proof $((f+g)+h)(u)=(f+g)(u)+h(u)=f(u)+g(u)+h(u)=(f+(g+h))(u)$
Abelian: $(f+g)(u)=f(u)+g(u)=g(u)+f(u)$ IMPORTANT: This uses $H$ abelian.
Inverses: Let $f \in Hom(G,H)$ then the only candidate to be inverse that could work, is the entrywise inverse $-f$ defined by $-f(x)=f(-x)$ . So let's check it does the job first:
$(f+(-f))(x)=f(x)+(-f)(x)=f(x)+f(-x)=f(x-x)=f(0)=0$ And the sum in the other direction has the same proof (or you remember this group $A$ is abelian).
And $-f$ is a homomorphism: $-f(x+y)=f(-(x+y))=f(-y-x)=f(-y)+f(-x)=f(-x)+f(-y)
=-f(x)+-f(y)$ Note this also uses $H$ to be abelian.
Best Answer
$2$.let $g\in G$ be an arbitrary element and $o(g)=l$. So there are $x,y\in\mathbb Z$ such that $2x+ly=1$. Thus $g=g^{2x}$.
$3$.As you know, the composition of two automorphisms are an automorphism.