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 $\Pr(T|S)=\frac{1}{2}$.
Let $E$ be the expected number of tosses it will take to get you back to even (so $E$ is the desired result).
Consider the first toss (not counting the initial $T$). If it is an $H$ you are done (in one toss).
Suppose your first throw (not counting the initial $T$) is again a $T$. Then you expect it will take you $E$ tosses to get back to where the game started, and another $E$ tosses to get to even. Thus your expected number from that state is $2E$.
It follows that $$E=.7\times 1+.3\times (2E+1)\implies E=\frac 52$$
Best Answer
If the first comes down heads, the expected value is $3$ since you get $HT$ the next time you flip tails, which will take an average of two more flips. If it comes up tails, then you start over. So the recursion can be written $$T = 3/2 + (T+1)/2 $$ which has the solution $T=4.$
The problem with yours is the middle term $(T+2)/4.$ You don't have to start all the way over in the case of $HH,$ you could get $HHT.$