Past exam of the graph Theory and Probabilistic Combinatorics course of the University of Oxford / 2009 / A / 1 / c

False. The path graph on four vertices is bipartite and its first and last edges form a perfect matching. It has no cycle at all, so it has no Hamiltonian cycle. This counterexample is connected as well.

True. Every cycle in a bipartite graph has even length because its vertices alternate between the two parts. On a Hamiltonian cycle, choose every other edge. These edges are disjoint and cover every vertex exactly once, so they form a perfect matching.

False. The complete bipartite graph $K_{2009,2010}$ is $2009$-connected. Deleting fewer than $2009$ vertices leaves at least one vertex in each part and therefore a connected complete bipartite graph; deleting the smaller part disconnects the larger independent set. But its two parts have unequal sizes, so a matching cannot cover them both. Equivalently it has an odd number of vertices. Thus it has no perfect matching.