if n=1 the meat will be eaten and the lion will fall asleep but wont get eaten by other lion.
if n=2 the meat wont get eaten (if one lion will eat the meat he will get eaten by the second lion)
if n=3 the meat will be eaten since the lion that will eat it knows that after he will eat it n will be equal to 2 and in that case he wont get eaten.
if n=4 the meat wont get eaten (like n=2)
and so on...Its easy to conclude that if n is even the meat wont get eaten and if n is odd the meat will get eaten.
I'll be nit-picky and say that only happens if the lions know the other lions are intelligent, which was not stated (so I assume to be false).
Is that something you're supposed to ask in the interview if they don't say?
For the pirate problem, Pirate 2 won't ever vote for anything that gives him less than 500, P3 won't vote for anything that gives him less than 499, P4 won't vote for anything that gives him less than 499, and P5 wants to live, so I say:
P5 proposes a split of P5 gets 0 P3 or P4 gets 499, P2 gets 0, and P1 gets 1. It's accepted, everyone lives.
I looked at it like this...
If it gets down to P2, the split is P2 gets 500, P1 gets 0. P2 votes for it, it passes.
If it gets down to P3, the split is P3 gets 499, P2 gets 0, P1 gets 1. P3 and P1 vote for it, it passes.
If it gets down to P4, the split is P4 gets 499, P3 gets 0, P2 gets 0, P1 gets 1. P4 and P1 vote for it, it passes. (P4 could also give the 1 to P2, doesn't matter)
P5 must satisfy two people other than himself, and the only way to do that is to give it all up. Since he wants to live, he does give it all up.