If there are 5 cards on the board and 3 of them are hearts and I don't have any hearts in my hand, what is the probability of someone else NOT having 2 hearts in their hand?
This is quite difficult to answer without making some assumptions. You could just calculate the chances of your opponent having a hand with a maximum of one heart based on the known and unknown cards, which comes to about 95%*.
However, this is a flawed approach because it assumes that your opponent is equally likely to get to the river with any two cards they are dealt. To truly answer your question, you'd have to model the likely range of hands which your opponent might get to the river with, which would require knowledge of the betting action, player tendencies, table image and other factors and is not an exact science.
*Calculated based on the chance of your opponent having two hearts being:
(10/45)*(9/44) = 0.045 (4.5%)
Therefore, the chance of them not having two hearts is the inverse:
1 - 0.045 = 0.955 (~95%)
combin(10,2) / combin(45,2) = 0.0454545455 = 1/22
1 - 1/22 = 21 / 22 = 0.9545454545