Given a directed acyclic graph (DAG) - what can you say about each case?
A. Let's prove that $v$ will be a vertex which can be reached by every node (it is the mother vertex of the transpose of the graph) by contradiction. Suppose that there is a node $u$ such that $v$ is unreachable by it. Consider every node in the DAG reachable by $u$. There must be at least one node in this set that has no leaving edges. (Otherwise you could start from $u$ and always follow a leaving edge from the current node so after at most $|V| - 1$ steps you would visit an already visited node. But this implies a cycle which isn't allowd in a DAG.) So no other node is reachable by $u$ and $u$ could be last in a topological cue instead of $v$ and we reached a contradiction. As a result $v$ is reachable by every node. This is equivalent with (A)'s statement as a node being reachable by every node can appear only last in any topological cue.

Also $v$ is a sink because if it wasn't there would be at least one edge leaving it so it couldn't be at the end of any topological cue. (A)'s statement implies that $v$ is a sink but the reverse doesn't always hold.

B. $v$ is a mother vertex (every node is reachable by it). The proof is similar.