Let $R=\mathbb{Z}[x]$, $I=(3,x) \subset R$. I want to calculate Ext$_R^1(R/I,R)$. I'll start with the short exact sequence, $$0 \to I \to R \to R/I \to 0 $$ where the maps are the obvious choices. Applying the $Hom_R(-,R)$ functor, we get $$0 \to Hom_R(R/I,R) \to Hom_R(R,R) \to Hom_R(I,R) \to Ext_R^1(R/I,R). $$ So, I want to find the co-kernel of the map $Hom_R(R,R) \to Hom_R(I,R)$, but I have no idea how to begin approaching this. I know that $Hom_R(R,R) \cong R$ as $R$-modules, and my typical strategy would be to find some isomorphism for $Hom_R(I,R)$, however, I'm not confident in my attempts at doing this.
It seems to me that we can create a bijection between $I$ and $R^2$ under the map $(f,g) \mapsto 3\cdot f+x \cdot g$, but I'm very dubious to call this an isomorphism. Indeed, $R/I\cong \mathbb{Z}/3\mathbb{Z}$. If $I \cong R^2$, then we'd have $R/R^2 \cong \mathbb{Z}/3\mathbb{Z}$. I'm not even really sure how to interpret the ring $R/R^2$, much less do I believe it's isomorphic to $\mathbb{Z}/3\mathbb{Z}$.
And so I go back to failing to understand how to work with $Hom_R(I,R)$, much less being able to approach a strategy of computing the Ext group.
Let $R$ be any commutative ring and let $a,b \in R$ be two elements. Let $I := \langle a,b \rangle$. Notice that $ba = ab$ is a relation between the generators of $I$. It follows that if $M$ is any $R$-module, then the map $$\hom(I,M) \to \{(m,n) \in M^2 \mid bm = an \},\quad \varphi \mapsto (\varphi(a),\varphi(b))$$ is well-defined and injective. It is also $R$-linear, but we don't need that here.
Now assume that $R$ is an UFD and $a,b$ are coprime*. Then the map above is bijective. The inverse map takes $(m,n)$ to the homomorphism $\varphi : I \to M$ defined by $\varphi(ra + sb) := rm + sn$. The main task is to prove that $\varphi$ is indeed well-defined $(\ast)$, but that's quite easy with the lemma $a \mid sb \implies a \mid s$ for $s \in R$.
Notice that bijectivity of the map can also be interpreted as the statement that
$0 \to R \xrightarrow{\begin{pmatrix} b \\ -a \end{pmatrix}} R^2 \xrightarrow{\begin{pmatrix} a & b \end{pmatrix}} I \to 0$
is an exact sequence. This is a free resolution of $I$, therefore useful for several computations (not here, though).
For $M := R$ we see that $\hom(I,R)$ consists of those $(m,n) \in R^2$ with $bm = an$. (I take this as an identification now. Notice that the inclusion $I \hookrightarrow R$ corresponds to the pair $(a,b)$.) Since $a,b$ coprime, we must have $ (m,n) = u (a,b)$ for some $u \in R$ (this is what you need to prove in $(\ast)$ above). This shows $R \to \hom(I,R)$, $1 \mapsto (a,b)$ is an isomorphism. The homomorphism $R \cong \hom(R,R) \to \hom(I,R)$ maps $1 \mapsto (a,b)$. Thus, $\hom(R,R) \to \hom(I,R)$ is in fact an isomorphism.
From the exact sequence
$\hom(R,R) \to \hom(I,R) \to \mathrm{Ext}^1(R/I,R) \to \mathrm{Ext}^2(R,R) = 0$
we thus get $\mathrm{Ext}^1(R/I,R) = 0$.
A natural question is what happens when $a,b$ are not coprime, but $R$ is still an UFD. The cases $a=0$ or $b=0$ are rather boring, so assume $a,b \neq 0$. Then $d := \mathrm{gcd}(a,b) \neq 0$, and $\frac{b}{d} \cdot a = \frac{a}{d} \cdot b$ is a relation in $I$ with coprime $\frac{a}{d},\frac{b}{d}$. It is again "the only" one: we have $\hom(I,M) \cong \{(m,n) \in M^2 : \frac{b}{d} \cdot m = \frac{a}{d} \cdot n\}$, and $R \cong \hom(R,R) \to \hom(I,R) \cong R \langle (\frac{a}{d},\frac{b}{d}) \rangle$ identifies with $ 1 \mapsto (a,b)$. Hence, $\mathrm{Ext}^1(R/I,R) \cong R/dR$.
*Of course, the special case $R = \mathbb{Z}[X]$, $a = 3$, $b = X$ is a pure distraction. Nothing will be easier in this special case.