Show that $\operatorname{Hom}_R(M, -)$ is a functor from the category of $R$-modules to the category of abelian groups. 
Show that $\operatorname{Hom}_R(M, -)$ is a functor from the category of $R$-modules to the category of abelian groups. 

Let $F$ be the functor.
Every $R$-module is also an additive abelian group, so we can send every $R$-modules $M$ to itself viewed as an additive abelian group (let's dentoe it as $M^*$). Every $R$-module homomorphism is also a group homomorphism (when $M$ is viewed as an additive abelian group). Let's denote $R$-module homomorphisms by $f$ and the corresponding group homomorphisms by $f^*$. For $f,g \in\operatorname{Hom}_R(M, -)$, we have 
$$F(f+g) = f^* + g^* = F(f) + F(g)$$
and 
$$F(1_M) = 1_M^*$$
where $1_M$ and $1_M^*$ denote the identities of $M$ and $M^*$, respectively. 
Do you think my answer is correct? 
Thanks in advance 
 A: Here is a little bit more conceptual answer.
The following fact is important: If $\mathcal{C}$ is an arbitrary category, and $M \in \mathcal{C}$, then there is a covariant Hom functor $\mathrm{Hom}(M,-) : \mathcal{C} \to \mathsf{Set}$. It maps an object $N$ of $\mathcal{C}$ to the set of morphisms $M \to N$, and a morphism $N \to N'$ is mapped to the map which takes $M \to N$ to the composition $M \to N \to N'$. It is easily checked that this is a functor. In fact, this is a direct consequence of the definitions of a functor and of a category.
Now let us look at $\mathcal{C} = \mathsf{Mod}(R)$ for some ring $R$. Then the Hom sets $\mathrm{Hom}(M,N)$ actually carry the structure of an abelian group, let's call this abelian group $\underline{\mathrm{Hom}}(M,N)$ (in order to distinguish it from its underlying set, which is not a group). The group operations are defined elementwise. We want to show that $\mathrm{Hom}(M,-) : \mathsf{Mod}(R) \to \mathsf{Set}$ lifts to a functor $\underline{\mathrm{Hom}}(M,-) : \mathsf{Mod}(R) \to \mathsf{Ab}$.
Since the forgetful functor $\mathsf{Ab} \to \mathsf{Set}$ is faithful (i.e. two homomorphisms of abelian groups equal iff their underlying maps of sets are equal), the defining properties of a functor are satisfied as soon as we show that for every $R$-linear map $f : N \to N'$, the induced map of sets $f_* : \mathrm{Hom}(M,N) \to \mathrm{Hom}(M,N')$ is actually additive and therefore lifts to a homomorphism of abelian groups $\underline{\mathrm{Hom}}(M,N) \to \underline{\mathrm{Hom}}(M,N')$. But this is an easy calculation just using the definitions and the assumption that $f$ is additive: For $g,h \in \mathrm{Hom}(M,N)$ and $m \in M$ we have
$f_*(g+h)(m)=f((g+h)(m))=f(g(m)+h(m))=f(g(m))+f(h(m))$
$=(f_* g)(m)+(f_* h)(m) = (f_* g + f_* h)(m).$
A: There are a few problems here. A small one first - $M^*$ and $f^*$ already mean (lots of) other things, so this notation is a little confusing. I think for this argument there's no real reason to distinguish between $M$ as an $R$-module or as an abelian group, and similarly for the morphisms, so I'd just continue calling them $M$ and $f$.
Now the major things - $\operatorname{Hom}_R(M,-)$ isn't a set, so it doesn't make sense to refer to $f,g$ in it. Furthermore, $F$ doesn't send functions of modules to the (equal) maps of abelian groups.
Continuing to denote the functor by $F$, we have that $F(N)=\operatorname{Hom}_R(M,N)$ (so $F$ does not just take $N$ to the underlying abelian group). You also need to define $F$ on morphisms; if $f\colon N_1\to N_2$ is a morphism of $R$-modules, then $F(f)\colon\operatorname{Hom}_R(M,N_1)\to\operatorname{Hom}_R(M,N_2)$ is a map of abelian groups; you should define $F(f)$ yourself - I can fill in details if you need, but it's good to work this out for yourself if you can. (The map $F(f)$ would often be denoted $f_*$ - if $G=\operatorname{Hom}_R(-,M)$, then $G(f)$ would often be called $f^*$, hence my earlier quibble with the notation.)
Then you need to check that $F(1_N)=1_{\operatorname{Hom}_R(M,N)}$, and that if $f\colon N_1\to N_2$ and $g\colon N_2\to N_3$, then $F(g\circ f)=F(g)\circ F(f)$.
As these are additive categories, you could also check that $F(f+g)=F(f)+F(g)$ for $f,g\colon N_1\to N_2$, so that $F$ is an additive functor, but this is more than you are asked.
