2 Flushes in Hold Em?

There are 10 different cases where both of your hole cards are suited and nobody else has two cards of the same suit as yours:
1. Nobody else has a card of the suit
2. One other player has one card of that suit
3. Two other players have one card of that suit
...
10. All nine other players have one card of that suit

Note that (A)C(B) = combin(A,B)

Case 10 has (39)C(2) x (37)C(2) x ... x (23)C(2) possible cases
For case 1, there are (9)C(1) = 9 possible hands that have the one card of that suit, 11 cards of that suit, 39 cards not of that suit that can be the other card, and (38)C(2) x (36)C(2) x ... x (24)C(2) possibilities for the other eight hands
Note this equals the case 10 value x 9 / 1 x 11 / (22 x 2)
For case 2, there are (9)C(2) = 36 possible hands that have one card of that suit, 11 x 39 possible "first hand" pairs, 10 x 38 possible "second hand" pairs, and (37)C(2) x (35)C(2) x ... x (25)C(2) possibilities for the other seven hands
This equals the case 9 value x 8 / 2 x 10 / (23 x 2)
Similarly, for cases 3 through 9, the case N value = the case (N - 1) value x (10 - N) / N x (12 - N) / ((21 + N) x 2).
Add these up, and divide by the total number of deals of the other 9 players, which is (50)C(2) x (48)C(2) x ... x (32)C(2), and you get a probability of 64.51% that no one else has 2 cards of your suit, which means the probability that at least one other player does is (100 - 64.51) = 35.49%.

Note that this is the probability that no one else has 2 cards of your suit after assuming that you have 2 suited cards.
In other words, if you have two suited cards, there is a 35.49% chance that at least one other player has 2 cards of the same suit as yours as well.
This does not take into account the probability of you having two suited cards in the first place.

Yes, I did run a simulation to check that value this time...

Hiding the answers in case others want to work them out I suppose? =P

So for part 2 you're asking what effect someone else having your suited cards would have on you drawing at a flush draw? Pretty simple... There would be 2 less outs for each of you to hit your flush (not caring whose is higher). Say you have 2 suited cards and the player next to you had 2 suited cards... ignoring if other players at the table have your suit or not (like BJ treating unseen cards as undealt cards), then you'd simply have 2 less outs to hit.

Typically when you have 2 suited cards and don't know anyone else's cards you assume you have 9 outs (2 in your had, 2 somewhere on the board, and 9 remaining of that suit in the deck). Going to the river there are 4 community cards, and your 2 hole cards out of the deck, thus you have 9/(52-6) chance of hitting your flush... 9/46 =19.57% chance.

IF someone else also had 2 of your suit (again, regardless of whose is higher) then that would simply take 2 outs away form both of you... Now your chance of catching the flush on the river would be 7/44 = 15.91% chance to hit a flush. So the net change of someone else having 2 of the same suit is "about" 3.5%.