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.
A simple way to find the probability is to condition on the result of the first round. It is clear there is some probability $p$ of obtaining a head before (though not necessarily immediately before) a $1$ or $2$. Call that the probability of winning.
We win if (i) we get a head on the first round or (ii) we get a tail, don't roll a $1$ or $2$, but ultimately win.
The probability of (i) is $\frac{1}{2}$.
For (ii), note that the probability of tail and then something other than $1$ or $2$ is $\frac{1}{2}\cdot \frac{4}{6}$. Given this has happened, the probability of ultimately winning is $p$. Thus
$$p=\frac{1}{2}+p\cdot \frac{1}{2}\cdot \frac{4}{6}.$$
Solve this linear equation for $p$. We get $p=\frac{3}{4}$.
Best Answer
The Geometric distribution can be described in two ways:
Since it is possible to get zero heads before the first tail, the support set is $k=0,1,\cdots$ in this case, and the pmf is $$P(X=k)=(1-p)^kp$$ with $p$ the probability of success (tails).
The cdf is $P(X\le k)$, which means the probability that the number of failures is less than or equal to $k$. The complementary event is $k+1$ times failure (heads). Hence the cdf is $$P(X\le k)=1-(1-p)^{k+1}$$