I agree that L1 is undecidable, because there is an easy reduction from it to the Halting Problem, which I assume you've already figured out. L2 is also clearly decidable; you just design a machine which takes the encoding of M as an input and counts the number of states. I think there is another easy reduction to show L1 intersect L2 is undecidable - essentially the same reduction as for L1 from the Halting problem, except that you also add 100 redundant states in order to satisfy L2. What do you think?
Edit: I misread less than 100 states for more than 100 states. In this case, we still use the reduction for L1 combined with a simulation to ensure using less than 100 states. Sorry, if I've confused you.