= Solution
The graph has edges $1\to2\leftarrow3$, $1\to4$, $2\to4$, $3\to5$, $4\to5$, $4\to6$, and $6\to5$. Any <D-separating set> for 1 and 6 must contain 4 because of the directed path $1\to4\to6$. Conditioning on 4 activates the collider $1\to4\leftarrow2$ and, through its descendant, the collider $1\to2\leftarrow3$. The remaining route through $3\to5\leftarrow6$ is open exactly when 5 is conditioned on and 3 is not. Hence all separating sets, among the nonendpoint vertices, are
$$
\{4\},\quad\{2,4\},\quad\{3,4\},\quad\{2,3,4\},
\quad\{3,4,5\},\quad\{2,3,4,5\}.
$$
For the second graph, $2\perp\!\!\!\perp5\mid1,6$ forces colliders on the unblocked two-edge paths: $2\to3\leftarrow5$ and $2\to4\leftarrow5$. Acyclicity then forces $1\to4$ and $6\to4$. Thus its edges are
$$
1\to2,quad2\to3,quad5\to3,quad1\to4,quad2\to4,
\quad5\to4,quad6\to4,quad6\to5.
$$
Solved by gpt-5.6-sol high.
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