The first step is correct, but the second step is not: You cannot say anything about the kernel after tensoring. Also note that your second step is purely formal and would apply to every additive functor which preserves epis. But not every such functor is right exact.
Let $M_1 \to M_2 \to M_3 \to 0$ be an exact sequence. We want to show that, for every module $N$, the sequence $M_1 \otimes N \to M_2 \otimes N \to M_3 \otimes N \to 0$ is exact, i.e. that $M_2 \otimes N \to M_3 \otimes N$ is a cokernel of $M_1 \otimes N \to M_2 \otimes N$. This means, by the universal property of the cokernel, that for every "test" module $T$, the sequence $0 \to \hom(M_3 \otimes N,T) \to \hom(M_2 \otimes N,T) \to \hom(M_1 \otimes N,T)$ is exact (as abelian groups, but then also as modules). By definition of the tensor product, this sequence is isomorphic to the sequence $0 \to \mathrm{Bilin}(M_3,N;T) \to \mathrm{Bilin}(M_2,N;T) \to \mathrm{Bilin}(M_1,N;T)$. $(\star)$
Thus, the claim is actually equivalent to a statement about bilinear maps. And this can be checked now directly. I will leave out the trivial steps. For the only interesting one, let $\beta : M_2 \times N \to T$ be a bilinear map which vanishes on $M_1 \times N$. Define $\gamma : M_3 \times N \to T$ as follows: If $m_3 \in M_3$, $n \in N$, choose a preimage $m_2 \in M_2$ of $m_3$ and define $\gamma(m_3,n):=\beta(m_2,n)$. This is well-defined, because every other choice of $m_2$ is of the form $m_2+x$ for some $x$ coming from $M_1$, and then $\beta(m_2+x,n)=\beta(m_2,n)+\beta(x,n)=\beta(m_2,n)$. One sees directly that $\gamma$ is bilinear because $\beta$ is. And course $\gamma$ is the desired preimage in $\mathrm{Bilin}(M_3,N;T)$.
This is not the most conceptual proof. You have already mentioned the one using adjoint functors. But we can also choose an alternative ending for the proof above: The sequence $(\star)$ is isomorphic to $0 \to \hom(N,\hom(M_3,T)) \to \hom(N,\hom(M_2,T)) \to \hom(N,\hom(M_1,T))$, which is exact because $\hom(N,-)$ is left exact and $\hom(-,T)$ is right exact.
And yet another ending (which explains Qiaochu's comment): The isomorphism $\mathrm{Bilin}(-,N;T) \cong \hom(-,\hom(N,T))$ shows that this functor is representable and therefore right exact, hence $(\star)$ is exact.
Neither is true for $R=\mathbb{Z}$ (i.e., for Hopfian abelian groups).
For (1), Corner gave an example of two Hopfian abelian grops whose direct sum is not Hopfian in
Corner, A. L. S., Three examples on Hopficity in torsion-free Abelian groups, Acta Math. Acad. Sci. Hung. 16, 303-310 (1965). ZBL0145.03302.
(2) is easier. $\mathbb{Q}$ is a Hopfian abelian group, but has a quotient $\mathbb{Q}/\mathbb{Z}$ which is not Hopfian, since multiplication by $n$ is surjective but not injective for any $n>1$.
Best Answer
You've already been given a method for surjectivity, so I'll try to explain a different general method for this problem which should help explain part (b).
As the other answer explained, a map out of $R^n$ is the same as defining where the basis vectors $e_i$ go. In other words, there is an isomorphism $Hom_R(R^n, M) \cong M^n$ sending $f \mapsto (f(e_i))_i$. This is, in fact, "natural". In other words, the following diagram commutes for any map $f: M \rightarrow N$.
where the vertical maps are the isomorphisms I described before.
Sorry for the poor typesetting, I'll try to improve it when I can.Anyway, this tells us that the sequence you gave is isomorphic to
$$ 0 \rightarrow M_1^n \xrightarrow{f^n} M_2^n \xrightarrow{g^n} M_3^n \rightarrow 0. $$
Exactness of this sequence is more easily checked.
For part (b), one would hope to do the same thing. However, it's not in general true that $Hom_R(R^\Delta, M) \cong M^\Delta$ naturally for $\Delta$ infinite. We'd have to replace $R^\Delta$ with $\bigoplus_{d \in \Delta} R$, the free $R$-module on $\Delta$. For ease of notation I'll call this $R^{\oplus \Delta}$. This is defined as $\{a \in R^\Delta : \text{ all but finitely many } a_d = 0\}$. Observe that this is just $R^\Delta$ for $\Delta$ finite. It turns out $Hom_R(R^{\oplus \Delta}, M) \cong M^\Delta$ naturally. You can then apply similar reasoning to prove this.
To get to the question you asked, the answer is that the sequence need not be exact. It will always be exact in the places you've already checked, but surjectivity can fail. The key notion here is projective modules. An $R$ module $P$ is said to be projective if $Hom_R(P, -)$ preserves exact sequences. It turns out that all modules preserve the left part of the exact sequence, so projectivity is equivalent to preserving the surjectivity part. The discussion above, by the way, leads to the result that arbitrary direct sums of $R$ are projective. However, it is not the case that all products of $R$ are projective. In fact, $\mathbb Z^{\mathbb N}$ is not projective. I'm not sure of an easy way to prove this, but as $\mathbb Z$ is a PID, it suffices to show that $\mathbb Z^{\mathbb N}$ is not free (i.e. isomorphic to a direct sum of copies of $\mathbb Z$). This isn't the easiest result, and is proven here.