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 $g^*$. $\blacksquare$
There are several parts to computing this, since one should first understand both constructions $\mathrm{Hom}_A(M,A)$ and $DN$, which are best thought of in terms of $A$-modules, and then we want to turn $D\mathrm{Hom}_A(M,A)$ back into a quiver representation. I will go through these constructions in general, before answering your specific problem at the end.
If $M$ is a left $A$-module, then the vector space $\mathrm{Hom}_A(M,A)$ becomes a right $A$-module via
$$ fa \colon M\to A, \quad m\mapsto f(m)a. $$
Let $e_i$ be the idempotent corresponding to vertex $i$. Then $Ae_i=P_i$ is the indecomposable projective $A$-module corresponding to vertex $i$, and
$$ \mathrm{Hom}_A(M,A)e_i = \mathrm{Hom}_A(M,Ae_i) = \mathrm{Hom}_A(M,P_i). $$
If $\alpha\colon i\to j$ is an arrow, then this yields an $A$-module map
$$ \alpha^\ast \colon P_j \to P_i, \quad a \mapsto a\alpha \quad\textrm{for $a$ in }P_j=Ae_j. $$
We therefore get an induced linear map
$$ \alpha^\ast \colon \mathrm{Hom}_A(M,P_j) \to \mathrm{Hom}_A(M,P_i), \quad f\mapsto\alpha^\ast f, \textrm{ which sends $m$ to }f(m)\alpha. $$
We next consider the vector space duality $D$. If $N$ is a right $A$-module, then $DN$ is a left $A$-module via
$$ a\xi\colon N\to k, \quad n\mapsto\xi(na). $$
In particular, $e_iDN=D(Ne_i)$. For an arrow $\alpha\colon i\to j$ we have the linear map $Ne_j\to Ne_i$, $n\mapsto n\alpha$, and then the induced linear map
$$ \alpha \colon D(Ne_i) \to D(Ne_j), \quad \xi\mapsto\alpha\xi, \textrm{ which sends $n\in Ne_j$ to $\xi(n\alpha)$.} $$
(Note that this is where the arrows get reversed: $\alpha\colon i\to j$ yields a map $Ne_j\to Ne_i$ for a right $A$-module $N$.)
We now put this together. For a left $A$-module $M$, the quiver representation corresponding to $D\mathrm{Hom}_A(M,A)$ has the vector space $D\mathrm{Hom}_A(M,P_i)$ at vertex $i$, and for an arrow $\alpha\colon i\to j$, we have the linear map
$$ \alpha\colon D\mathrm{Hom}_A(M,P_i) \to D\mathrm{Hom}_A(M,P_j), \quad \xi\mapsto\alpha\xi, \textrm{ which sends $g\in\mathrm{Hom}_A(M,P_j)$ to $\xi(\alpha^\ast g)$}. $$
OK. So what does this do in your special case, where $M=P_2$? We have $\mathrm{Hom}_A(M,P_2)=k\,\mathrm{id}$, and $\mathrm{Hom}_A(M,P_1)=k\,\alpha^\ast$, where $\mathrm{id}$ is the identity map and $\alpha\colon 1\to 2$ is the arrow.
Let $\xi$ and $\eta$ be the respective dual bases, so that $D\mathrm{Hom}_A(M,A)$ is one dimensional at each vertex, with basis $\eta$ at vertex 1 and $\xi$ at vertex 2.
Finally, $\alpha\eta(\mathrm{id})=\eta(\alpha^\ast)=1=\xi(\mathrm{id})$, so $\alpha$ sends $\eta$ to $\xi$, and we have the quiver representation $k \xrightarrow{1} k$ as expected.
Best Answer
The functor $\nu$ is additive and sends $A$ to the injective module $DA$, so sends finite dimensional projectives to finite dimensional injectives.
Similarly the functor $\nu^-$ is additive and sends the injective cogenerator $DA$ to $$ \mathrm{Hom}_A(DA,DA) \cong \mathrm{Hom}_k(A\otimes_ADA,k) \cong D^2A \cong A, $$ so send finite dimensional injectives to finite dimensional projectives.
To see that the functors form an adjoint pair on these subcategories, we can show that there is a natural isomorphism of bifunctors $$ \mathrm{Hom}(\nu^-I,P) \cong \mathrm{Hom}(I,\nu P) $$ with $P$ finite dimensional projective and $I$ finite dimensional injective.
We have natural isomorphisms $$ \mathrm{Hom}(I,D\mathrm{Hom}(P,A)) \cong D(\mathrm{Hom}(P,A)\otimes_AI) \cong D\mathrm{Hom}(P,I) $$ where the last follows since $P$ is finite dimensional projective (dual basis Lemma).
Next, for any two finite dimensional (right and left) modules $X$ and $Y$ we have $$ \mathrm{Hom}_A(X,DY) \cong D(X\otimes_AY) \cong \mathrm{Hom}_A(Y,DX). $$
We now get natural isomorphisms $$ \mathrm{Hom}(\mathrm{Hom}(DA,I),P) \cong \mathrm{Hom}(\mathrm{Hom}(DI,A),P) \cong D(DP\otimes_A\mathrm{Hom}(DI,A)) \cong D\mathrm{Hom}(DI,DP). $$ The last uses the dual basis Lemma for the finite dimensional projective right module $DI$. Applying the observation above once more, this is isomorphic to $D\mathrm{Hom}(P,I)$.
Now that we have adjointness, we have the unit and count, and the computations we did at the beginning show that these are isomorphisms for $DA$ and $A$ respectively. Since they are additive, they are isomorphisms for all (finite dimensional) injectives and projectives, and so we have the desired (adjoint) equivalence between these subcategories.