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?
In the discrete case, the support of the random variable $X$ is the set of values of $x$ such that $\Pr(X=x)\ne 0$. In our case, the possible values of $X$ range from $1$ to $7$, so the support of $X$ is $\{1,2,3,4,5,6,7\}$.
Added: We find the distribution of $X$, by specifying $\Pr(X=x)$ for all values $x$ in the support of $X$.
In order for $X$ to be $1$, we need to roll a $1$ and toss a tail. The probability of this is $\frac {1}{6}\cdot \frac{1}{2}$. Thus $\Pr(X=1)=\frac {1}{12}$.
The random variable $X$ can be $2$ in two ways: (i) we get a $2$ on the die, and roll a tail or (ii) we roll a $1$ on the die, and toss a head. The probability of (i) is $\frac{1}{6}\cdot\frac{1}{2}$. The probability of (ii) is the same. It follows that $\Pr(X=2)=\frac{1}{6}$.
You can handle the probabilities that $X=3$, $X=4$, and so on to $7$.
For the cdf $F_X(x)$, recall that $F_X(x)$ is the probability that $X\le x$, and is defined for all real $x$.
If $x\lt 1$, the $\Pr(X\le x)=0$, so $F_X(x)=0$.
If $1\le x\lt 2$, then $\Pr(X\le x)=\frac{1}{12}$, so in this interval $F_X(x)=\frac{1}{12}$.
If $2\le x\lt 3$, then $\Pr(X\le x)=\frac{1}{12}+\frac{1}{6}$. Thus $F_X(x)=\frac{3}{12}$ in this interval.
Continue. Don't forget about $F_X(x)=1$ if $x\ge 7$.
The remaining questions will probably not cause any difficulty.
Best Answer
Hint:
$$E(T)=a_1E(X_1)+a_2E(X_2)+\ldots+a_nE(x_n)$$
$$\text{Var}(T)=(a_1)^2\text{Var}(X_1)+(a_2)^2\text{Var}(X_2)+\ldots+(a_n)^2\text{Var}(X_n)$$
Treat the $a_i$'s as constants. For example, for $T=4S-3H$, $$E(T)=E(4S)-E(3H)=4E(S)-3E(H)$$
$S$ follows the Bernoulli Distribution because our experiment of rolling a dice once has only two outcomes: either you roll a certain number (success) or you don't (failure). For a Bernoulli Distribution,
$$E(S)=p=\frac{1}{6}$$
Can you determine the distribution of $H$ and find $E(H)$ from here?