The two-tailed coin can't be the one that was chosen. So one of the two fair coins was chosen, and we know it landed heads. The fair coin that was not chosen could land either heads or tails, with equal probabilities.
So the possible values of $X$ are $1$ and $2$, each with probability $1/2$.
As for the conditional probability of the "third toss", the question is a bit
ambiguous. Is the coin that was chosen the first toss, or could the coins have been
tossed in any order?
Let $Y$ be the number of heads following the first tail. We calculate $\Pr(Y=0)$, $\Pr(Y=1)$, $\Pr(Y=2)$, and $\Pr(Y=3)$. It will be done by a patient examination of cases. (If we were dealing with say $10$ tosses, then a more general approach would be worthwhile.)
The easiest one is $\Pr(Y=3)$. For $Y=3$ can only happen if we get a tail and then $3$ heads. The probability of this is $\frac{1}{2^4}$.
Now let's go to the other end, and find $\Pr(Y=0)$. This can happen in several ways. We could get TTTT, probability $\frac{1}{2^4}$. Or we could get HTTT, probability $\frac{1}{2^4}$. Other possibilities are HHTT,HHHT, HHHH. Add up. We get $\frac{5}{2^4}$.
Next we find $\Pr(Y=1)$. Here we use some machinery, not much. We could have $T$, then exactly one H. This has probability $\frac{1}{2}\cdot \binom{3}{1}\cdot \frac{1}{2^3}$, that is, $\frac{3}{2^4}$. Or else we could get HT, then exactly $1$ H. The probability is $\frac{2}{2^4}$. Or else we could have HHTH, probability $\frac{1}{2^4}$. Add up.
Finally, for $Y=2$ we can do a similar calculation, or use the fact that probabilities sum to $1$. It is best to do both, as a partial check on the correctness of our calculations.
Best Answer
An alternate solution is as follows: imagine you flipped all the coins twice. Then any coin that gave you heads on the first flip or the second flip would be one of the ones you want to count. The probability of getting at least one head in two flips is $3/4$, so the expected number of coins that get at least one head is $10 \times 3/4 = 7.5$.