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 $\Pr(T|S)=\frac{1}{2}$.
As you say, the probability of a person getting one head and one tail is $\frac12$. The probability of getting two heads or two tails is $\frac12$. Altogether, we'll have:
$${3 \choose 2}\times\left(\frac12\right)^2\times\frac12 = \frac38.$$
The $\left(\frac12\right)^2$ accounts for the two people who get a head and a tail, and the final $\frac12$ accounts for the one person who doesn't.
Best Answer
I think it is worth stressing that you can think of the outcomes as ordered or not, as you wish. That does not change the problem, nor does it change the final answer, but it does alter the computation.
Ordered: There are $4$ possible outcomes $$HH, HT, TH, TT$$ each of which has probability $\frac 14$. By inspection, the winning outcomes are $HT, TH$ so the answer is $\frac 12$.
Unordered: Now there are $3$ possible outcomes $$\{H,H\}, \{H,T\}, \{T,T\}$$ but these three are not equi-probable. Indeed, there is only one way to get $\{H,H\}$, say, which makes the probability of that outcome $\frac 14$. Same for $\{T,T\}$. There are two ways to get $\{H, T\}$ so that has twice the probability, again yielding $\frac 12$.
This comes up for dice throws, for example. If you consider ordered outcomes, then each outcome has probability $\frac 1{36}$. If you consider them as unordered, then the probability of getting distinct values, like $\{1,2\}$ is $\frac 2{36}=\frac 1{18}$ while the probability of getting a double like $\{1,1\}$ is still $\frac 1{36}$. Personally, I try to stick to equi-probable cases wherever possible, as that simplifies the computations. But it's up to you, and in some situations you really can't avoid considering outcomes with differing probabilities.