a. We consider an arbitrary group $O$. Each of the rest $111$ groups has exactly one student belonging to $O$. Since $111 = 11 \cdot 10 + 1$, from the pigeonhole's rule we conclude that there exists one student $x \in O$ who belongs at least to $11$ other groups. Hence $x$ belongs at least to $12$ groups, say $O_1, O_2, \dots, O_{12}$.

b. We will prove that $x$ belongs to all groups. In fact, let $x \notin O'$. Then the group $O'$ has exactly one common student with the groups $O_1, O_2, \dots, O_{12}$. Since $O'$ has $11$ students, then there will exist two groups, say $O_i, O_j$ having the same common student with the group $O'$, say $y$. However, in this case the groups $O_i, O_j$ will have two common students $x$ and $y$, absurd.

Therefore $x$ belongs to all groups.