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 $\Pr(T|S)=\frac{1}{2}$.
If the coin is fair then the number of heads in a row before you flip has no affect on the probability of another head so the the answer for a fair coin is $\frac{1}{2}$.
If this is a coin with both heads then you clearly has no chances to see tail at all.
Using Bayes' theorem you can say that
$$P(\mbox{your coin is double-headed}) = \frac{1\cdot \frac{1}{2}}{1\cdot \frac{1}{2} + \frac{1}{8}\cdot \frac{1}{2}} = \frac{8}{9}$$
Just to add more generality, using the same approach you can show that the probability that the coin is double-headed after $N$ heads in a row and no tails is
$$P = \frac{p}{p+(1-p)\cdot2^{-N}},$$
where $p$ is the probability you think the coin is double-headed before you receive any information via flips. After some rather big $N$ it will be very close to $1$ for any reasonable $p$.
Best Answer
You had no prior information about which coin it was, so suppose they were equally likely.
Three coin tosses have come up heads. That would always happen with the second coin, but only happen once in eight with the first coin. So the second coin is now eight times more likely than the first coin.
Put the new probabilities of the two coins into your formula.