The friend says yes if he got one head or two.
When a coin is tossed twice, the following $4$ outcomes are equally likely: HH, HT, TH, TT. For $3$ of these outcomes, the friend says yes. In $2$ of these outcomes, there is a tail. So the probability that there is a tail given that there is at least one head is $\dfrac{2}{3}$.
This sort of semi-formal argument can be treacherous. So let's do it more formally. Let $B$ be the event there is at least $1$ head, and let $A$ be the event there is a tail. We want $\Pr(A|B)$ (the probability of $A$ given $B$).
By a standard formula,
$$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}.$$
The probability of $B$ is $\dfrac{3}{4}$.
The probability of $A\cap B$ is $\dfrac{2}{4}$. Divide.
Added: A second part has been added. This one requires interpretation. Answers to similar problems can be quite interpretation-dependent.
Suppose that with probability $\frac{1}{2}$ we get to have a peek at the result of toss $1$, and with probability $\frac{1}{2}$ we get to see the result of toss $2$. Let $S$ be the event we see a head, and let $T$ be the event there is a tail.
We want $\Pr(T|S)$. The computation is much like the one in the answer to the first question. After examination of cases, we find that $\Pr(S)=\frac{1}{2}$ and $\Pr(S\cap T)=\frac{1}{4}$. From that we conclude that $\Pr(T|S)=\frac{1}{2}$.
Best Answer
Take it one step at a time.
With probability $\frac12$ you toss a head and then roll one die; the probability that the die comes up $6$ is $\frac16$, so the overall probability of getting your total of $6$ in this way is $\frac12\cdot\frac16=\frac1{12}$.
With probability $\frac12$ you toss a head and then roll two dice. What’s the probability of getting a total of $6$ when you roll two dice? Half that is the probability of getting your total of $6$ in this way.
Now just add the two partial results to get the desired probability.