(2) and (3) really are easy though. Both can be solved by deriving counterexamples for the options until one is left.
For (2), imagine if there were only two teachers in the convention and they shook hands with each other once - that invalidates option C. If they shake again, that invalidates option B and E. A and D both talk about an even number of teachers, so let's imagine there were three teachers. If each teacher shakes hands with the other two once, that invalidates D. So A is the answer. I guess some form of graph theory is involved with solving it properly.
(3) can be reasoned in the same way by imagining quadrilaterals for each option. A quad with two adjacent tiny edges and two long edges symmetric along one axis (like a square with one vertice pulled away) gives a rectangle, which invalidates A, B and E. Making it asymmetric by moving the far vertex invalidates C. So only D is left.