Probability

P(X>=1) = 1-P(X=0)
P(X=0) = 15/10C3
P(X>=1) = 105/120

10a. 4C1*13C10*39C0 / 52C10
10b. 4C2*[13C1*13C9+13C2*13C8+...+13C9*13C1]*26C0 / 52C10

a. 1/4P4 = 1/24
b. if you have 3 matching, you must also have 4 matching, so same as (a)? is there overlap?
c. 1-(6/24)
d. 6/24

Actually, you're on the right track for this one but wrong. As you said if you have 3 matching the 4th must match too therefore P(X=3) = 0. You can never have exactly 3 pairs matching.

Working backwards:
EV of 1 roll = 7/2
For 2 rolls, keep roll on 4, 5, or 6, re-roll on 1, 2 or 3. EV is therefore:
1/2 * 5 + 1/2 * 7/2 = 17/4
For 3 rolls, keep roll on 5 or 6, re-roll on 1, 2, 3, or 4. EV is therefore:
2/3 * 4.25 + 1/3 * 5.5 = 14/3

I think that's incorrect because you should stop only when you roll is lower than your future potential EV.

EV of 1 roll = 7/2
So, technically, you should re-roll if you get 1, 2 or 3.

However, if you look at it from the other end, you still have two rolls left, which will yield an EV of 17/4, so I think you should actually stop rolling only if you get 5 or 6 on your first roll.

Um, you did the exact same thing I did.

Your first roll keeps 5s or 6s, meaning that the EV of all 1st rolls that are kept is 5.5. If you choose to re-roll, your second roll is kept on 4, 5, or 6, meaning the EV of all 2nd rolls that are kept is 5. If you roll a 3rd time, you have an EV of 3.5.
So the total EV is:

(1/3)*5.5 + (2/3)*(1/2)*5 + (2/3)*(1/2)*3.5 = 14/3