Difficulty: Hard

Since we can not place any more coins on the table, each point of it is at distance at most 2r from the center of some coin, where r is the radius of the coin. Now shrink the entire table twice in width and length, then replace every shrunk coin with a full sized one. This way the small table will be completely covered because every point of it will be at distance at most r from the center of some coin. Add three more of these smaller tables, covered with coins, to create a covering of the big table.

For every subset of 2 people you pick among the 5, there should be a lock which none of the 2 can unlock, and each of the remaining 3 people can unlock. Clearly, the lock in question cannot be the same for any two different subsets of 2 people you choose. Therefore the number of locks you need is at least the number of different 2-element subsets of a 5-element set, which is 5!/(2!3!)=10. This number is sufficient as well – just give keys to a different group of 3 people for every lock.

Yes, you can do this. The easiest way is to use generating functions. Using simple polynomial algebra, you can see that

(x + x₂ + x₃ + x₄ + x₅ + x₆)² = (x + 2x₂ + 2x₃ + x₄)(x + x₃ + x₄ + x₅ + x₆ + x₈).

Therefore, if you take a dice with spots {1, 2, 2, 3, 3, 4}, and a dice with spots {1, 3, 4, 5, 6, 8}, their sum will have the same probability distribution.

The zebras can survive forever. They choose 100 parallel strips with width 300m each, then start on points on their mid-lines. If the lion lands on some zebra’s strip, the zebra simply jumps 100m away from the lion, along its mid-line.