Zero element of Ext$^n$ groups In every Abelian category, given a non-negative integer $n$ and objects $A$ and $B$, one can define the (Yoneda) Ext-set $\text{Ext}^n(B,A)$. The construction is explained in this Wikipedia page. What is not explained, however, is how to make these sets into Abelian groups. The operation is defined, but what is the zero element? For $n=1$ there is an obvious choice of a trivial exact sequence, namely, the split exact sequence
$$0 \to A \to A \oplus B \to B \to 0$$
with the standard maps. But I wonder how a trivial exact sequence of the form
$$0 \to A \to ? \to ? \to B \to 0$$
should look like. And the same for higher $n$.
 A: If $A$ and $B$ are objects in the abelian category $\mathcal{A}$, the (equivalence class) of the exact sequence
$$0\to A\to A\oplus B\to B\to 0$$
is the zero object in $\text{Ext}^{1}(B,A)$.
For $n>1$, the zero object of $\text{Ext}^{n}(B,A)$ is the (equivalence class) of the sequence
$$0 \to A\to A \to 0 \to\cdots\to 0\to B\to B\to 0$$
where the maps $A\to A$ and $B\to B$ are the identity.
The group structure on both is a bit fiddly. Consider the maps $\Delta:A\to A\oplus A$ given by $\begin{pmatrix}1 \\ 1\end{pmatrix}$  and $\nabla:A\oplus A\to A$ given by $\begin{pmatrix}1 & 1\end{pmatrix}$. The operation $+$ on $\text{Ext}^{i}(B,A)$ is $$\textbf{E}_{1}+\textbf{E}_{2} := \nabla (\textbf{E}_{1}\oplus\textbf{E}_{2})\Delta$$
where the direct sum is what you expect. 
A proof that these form an abelian group can be found in Chapter 7 (called Extensions) of book Theory of Categories by Mitchell. It also shows that  these coincide with the usual derived functor definition of Ext groups. 
A: First one word of caution: for a general abelian category, even $\mathrm{Ext}^1(X,Y)$ can fail to be a set.
As people have pointed out above, the zero element in $\mathrm{Ext}^2(D,A)$ is represented by the four term exact sequence
$$ 0 \to A \xrightarrow{1} A \xrightarrow{0} D \xrightarrow{1} D \to 0. $$
That still leaves the question of whether a given four term sequence is zero. For this there is the following nice answer. Given an exact sequence
$$ \eta\colon 0 \to A \to B \to C \to D \to 0, $$
let $I$ be the image of the map $C\to D$. Then $\eta$ is zero in $\mathrm{Ext}^2(D,A)$ if and only if there exists an object $X$ fitting into an exact commutative diagram
$$ \require{AMScd} \begin{CD}
@.@. 0 @. 0\\
@.@. @VVV @VVV\\
0 @>>> A @>>> B @>>> I @>>> 0\\
@. @| @VVV @VVV\\
0 @>>> A @>>> X @>>> C @>>> 0\\
@.@. @VVV @VVV\\
@.@. D @= D\\
@.@. @VVV @VVV\\
@.@. 0 @. 0
\end{CD} $$
In fact, given another exact sequence
$$ \eta' \colon 0 \to A \to B' \to C' \to D \to 0 $$
we have that $\eta=\eta'$ in $\mathrm{Ext}^2(D,A)$ if and only if there exists an object $X$ fitting into the exact commutative diagram
$$ \require{AMScd} \begin{CD}
@. 0 @. 0 @. 0\\
@. @VVV @VVV @VVV\\
0 @>>> A @>>> B @>>> I @>>> 0\\
@. @VVV @VVV @VVV\\
0 @>>> B' @>>> X @>>> C @>>> 0\\
@. @VVV @VVV @VVV\\
0 @>>> I' @>>> C' @>>> D @>>> 0\\
@. @VVV @VVV @VVV\\
@. 0 @. 0 @. 0
\end{CD} $$
