At a minimum, there are only two things left to prove to complete the equivalences.
I'll prove (1) $\implies$ (3) and (4) $\implies$ (2), since you know (3) $\implies$ (4) and (2) $\implies$ (1).
I'm going to state two facts about $\newcommand\Ext{\operatorname{Ext}}\newcommand\Hom{\operatorname{Hom}}\Ext$, and if you're not familiar with them, then I'd suggest looking into these, because they're a bit beyond the scope of an answer to reprove here.
Fact 1 If $B\to I^0\to I^1\to \cdots \to I^n\to\cdots$ is any injective resolution of $B$, then for any $A$,
$$\Ext^n(A,B)\cong H^n(\Hom(A,I^\bullet))
$$
Fact 1 gives (1) $\implies$ (3), since $B\to 0 \to 0 \to 0 \to \cdots $ is already an injective resolution of $B$ when $B$ is injective, so
$$\Ext^n(A,B) \cong H^n(\Hom(A,B)\to 0 \to 0 \to 0 \to 0),$$
so $\Ext^i(A,B)=0$ for $i>0$ (and any $A$).
Fact 2 If $0\to A' \to A\to A''\to 0$ is any short exact sequence of $R$-modules, then there is a long exact sequence for $\Ext$ for any $B$:
$$
0\to \Hom(A'',B) \to \Hom(A,B)\to \Hom(A',B)\to \Ext^1(A'',B)\to \Ext^1(A,B)\to \cdots
$$
$$\Ext^n(A',B)\to\Ext^{n+1}(A'',B)\to \Ext^{n+1}(A,B)\to \Ext^{n+1}(A',B)\to \cdots
$$
Fact 2 gives (4) $\implies$ (2), since if $\Ext^1(A,B)=0$ for all $A$, then for any short
exact sequence $0\to A'\to A\to A''\to 0$, we have the long exact sequence
$$0\to \Hom(A'',B)\to \Hom(A,B)\to \Hom(A',B)\to \Ext^1(A'',B)=0,$$
since we assumed that $\Ext^1(A'',B)=0$ for any $A''$, so $\Hom(-,B)$ is an exact functor.
Edit:
Also I missed this when reading your question, but I realized I didn't directly address your first question about proving exactness in the middle when proving (1) $\implies$ (2).
This also follows from Fact 2 above, but that's overkill, there is actually an elementary proof that for any short exact sequence $\newcommand\toby\xrightarrow 0\to A'\toby{f} A\toby{g} A''\to 0$, and any $B$, the sequence
$$0\to \Hom(A'',B)\toby{g^*} \Hom(A,B)\toby{f^*} \Hom(A',B)$$
is exact. Then $B$ being injective is equivalent to the last map being surjective for all short exact sequences.
Proof
Since $gf=0$, we have $f^*g^*=(gf)^*=0$, which means $\newcommand\im{\operatorname{im}}\im g^*\subseteq \ker f^*$ so we have two things to prove:
(a) injectivity of $g^*$, and (b) that $\ker f^*\subseteq \im g^*$.
(a) If $\phi : A''\to B$ is some map, and $g^*\phi = \phi\circ g =0$, then,
if $x\in A''$ is any element, since $g$ is surjective, $x=g(a)$ for some $a\in A$, so $\phi(x) = \phi(g(a))=0$. Therefore $\phi=0$, so $g^*$ is injective.
(b) Suppose $\phi : A\to B$ is in the kernel of $f^*$, so $\phi\circ f =0$. Then since $f$ is injective and $g$ is surjective, we can regard $A'$ as a submodule of $A$ and we have that $A''\cong A/A'$. Then $\phi : A\to B$ is a morphism that is zero on $A'$, so we know that it induces a morphism $\phi' : A''\to B$ defined by $\phi'(g(a))=\phi(a)$. But this is exactly what it means to say $\phi = g^*\phi'$, so $\phi$ is in the image of $g^*$. $\blacksquare$
