= Solution
Choose a <projective presentation> $0\to K\xrightarrow{j}P\xrightarrow{\pi}C\to0$, with $P$ a <projective module>. Continuing it to a <projective resolution> shows
$$
\operatorname{Ext}_R^1(C,A)\cong\frac{\operatorname{Hom}_R(K,A)}{\{\ell j:\ell\in\operatorname{Hom}_R(P,A)\}}.
$$
Indeed a degree-one cocycle in the <Hom functor> applied to the <projective resolution> descends to $K$, while the degree-one coboundaries are exactly restrictions of maps from $P$. This establishes the description of the <Ext functor> without first assuming the extension classification.
For a <module extension> with inclusion $i:A\to B$ and quotient map $r:B\to C$, projectivity lifts $\pi$ to $s:P\to B$. Since $rsj=0$, there is a unique map $h:K\to A$ with $ih=sj$. A different lift changes $h$ by $\ell j$ for some $\ell:P\to A$. An <equivalence of module extensions>, which is the identity on both end modules, also preserves this class. We have therefore defined a map from extension classes to $\operatorname{Ext}_R^1(C,A)$.
Conversely, for $h:K\to A$, form the <pushout of a module extension>
$$
B_h=(A\oplus P)/\{(h(k),-j(k)):k\in K\}.
$$
The map $A\to B_h$ sends $a$ to $[a,0]$, and $B_h\to C$ sends $[a,p]$ to $\pi(p)$. The first is injective because $j$ is injective. If $\pi(p)=0$, write $p=j(k)$; then $[a,p]=[a+h(k),0]$, proving exactness in the middle. The last map is surjective. Thus $B_h$ is a <module extension>. If $h'=h+\ell j$, the map $[a,p]\mapsto[a-\ell(p),p]$ is an <equivalence of module extensions> from $B_h$ to $B_{h'}$. Finally $[a,p]\mapsto i(a)+s(p)$ identifies the <pushout of a module extension> built from an existing extension with its middle module. These two constructions are inverse, proving \b[the classification]:
$$
\boxed{\mathcal E(C,A)\ \cong\ \operatorname{Ext}_R^1(C,A).}
$$
The zero class corresponds to a <split short exact sequence>. Fixed end modules matter: equivalence does not permit arbitrary automorphisms of $A$ or $C$.
In the second calculation the acting group is the infinite cyclic group. Its modules are modules over the <group ring> $S=\mathbb Z[z,z^{-1}]$, not merely over the underlying ring $\mathbb Z$. The <trivial representation> has $z$ acting as the identity. Its <projective resolution> is
$$
0\longrightarrow S\xrightarrow{z-1}S\longrightarrow\mathbb Z\longrightarrow0.
$$
Multiplication by $z-1$ is injective, and the augmentation quotient is $\mathbb Z$. Applying the <Hom functor> into the trivial module makes the differential zero, so $\operatorname{Ext}_S^1(\mathbb Z,\mathbb Z)\cong\mathbb Z$. Taking two copies gives \b[$\boxed{\mathcal E(\mathbb Z^2,\mathbb Z)\cong\mathbb Z^2}$].
An explicit representative of this <extension of trivial modules for an infinite cyclic group> is the abelian group $B_{m,n}=\mathbb Zu\oplus\mathbb Zv\oplus\mathbb Zw$, with injection $1\mapsto u$, quotient $v\mapsto(1,0)$, $w\mapsto(0,1)$, and action
$$
\boxed{zu=u,\qquad zv=v+mu,\qquad zw=w+nu,\qquad(m,n)\in\mathbb Z^2.}
$$
This is an invertible action: the inverse subtracts the same multiples of $u$. Every underlying abelian-group extension splits because $\mathbb Z^2$ is free, so any group-module extension has this form after choosing lifts of the quotient basis. Replacing those lifts by multiples of $u$ does not alter $m,n$. An equivalence fixing the ends has exactly such changes of lifts, so two representatives are equivalent precisely when their ordered pairs agree. Only $(0,0)$ splits as a group-module extension; overlooking the group action would incorrectly give a single class.
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