= Solution
Definition 1 satisfies Property 1 but not Property 2. Let $S_1$ contain every non-descendant that blocks some backdoor path. Every backdoor path begins $A\leftarrow P$ for a parent $P$ of $A$; whenever that path can transmit confounding, $P\in S_1$. Conditioning on $S_1$ therefore blocks every such path at its first nonendpoint vertex, so $S_1$ is sufficient.
For failure of Property 2, consider the faithful graph with arrows
$$
C\to A,qquad C\to Y,qquad C\to X_i\to Y.
$$
The variable $X_i$ blocks the backdoor path $A\leftarrow C\to X_i\to Y$, so Definition 1 calls it a confounder. Every sufficient set containing $X_i$ must nevertheless contain $C$ to block $A\leftarrow C\to Y$. Once $C$ is included, deleting $X_i$ leaves a sufficient set. Thus $X_i$ can never be essential as Property 2 demands.
Definition 2 satisfies Property 2 but not Property 1. If $X_i$ belongs to every <minimal sufficient adjustment set>, choose one such set $I$ and put $J=I\setminus\{i\}$. By minimality, $J\cup\{i\}$ is sufficient and $J$ is not, proving Property 2.
For failure of Property 1, use the faithful chain-shaped backdoor path
$$
A\leftarrow C_1\to C_2\to Y.
$$
Both $\{C_1\}$ and $\{C_2\}$ are minimal sufficient adjustment sets. No variable belongs to every minimal sufficient set, so Definition 2 labels no variable a confounder, but the empty set is not sufficient.
Definition 3 satisfies Property 1 but not Property 2. Under faithfulness, its associational criterion contains enough non-descendants to block every open backdoor path. Indeed, if such a path remained open, its first unconditioned parent of $A$ would be D-connected to $A$ and, after a suitable conditioning set, to $Y$ given $A$; faithfulness would place that parent in the Definition 3 set, a contradiction. Thus adjusting for all variables selected by Definition 3 is sufficient.
For failure of Property 2, consider
$$
Z\to A\leftarrow U\to Y,qquad A\to Y,
$$
with both $Z$ and $U$ observed and a faithful distribution. The <instrumental variable> $Z$ is associated with $A$. Conditioning on the <collider> $A$ opens $Z\to A\leftarrow U\to Y$, so $Z$ is associated with $Y$ given $A$ and Definition 3 calls it a confounder. Any sufficient set containing $Z$ must also contain $U$ to block $A\leftarrow U\to Y$, but $\{U\}$ is already sufficient. Hence removing $Z$ never destroys sufficiency, violating Property 2.
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