When is splitting a collection coins two ways NP-complete?
Problem 2 is the same as the NP-complete problem PARTITION. Problem 3 is also NP-complete, since we can always arrange all denominations to be multiples of 10 cents.

The most interesting case is Problem 1. If the jar contains in total $2S$ cents, then we are looking for a subset of the jar summing to $S$. If $S$ is odd then there is no solution unless the jar contains a 1 cent coin, in which case we take one such coin and put it aside (it forms part of the solution), and update the target sum to $S-1$. Is $S$ is even we do nothing. In both cases, the new target sum is even. The number of 1 cent coins which appear in a solution must be even, and so we can group the 1 cent coins in pairs, replacing them with 2 cent coins (if there is an odd number, we just throws the odd one out). Now we repeat the same procedure with the 2 cent coins, the 4 cent coins, and so on, until we either reach the target sum or find out that the target sum is not reachable. This algorithm runs in polynomial time.