I don't really know how to solve this problem using the pigeonhole principle. My attempt so far was to show that you can choose 10 possible 3 element subsets and from those you can choose 3 possible 2 element subsets, making for 30 possible options of pigeons that cannot be together? However, this doesn't really make sense to me because there's only 2 holes. Please give me advice on how to start thinking about this problem.
First question: yes
$$(1, 2), (2, 3), (3, 4), (4, 5), (5, 1)$$ $$(1, 3), (3, 5), (5, 2), (2, 4), (4, 1)$$