There's a nice discussion of Penney's game in Section 8.4 of Concrete Mathematics. Using the techniques described there, the answers I get (confirming joriki's) are
(a) $(1 + p + p^2)q$, which is $0.657$ when $p = 0.7$, $q = 0.3$.
(b) $\frac{1-pq}{2-p}$, which is $\frac{79}{130} \approx 0.608$ when $p = 0.7, q = 0.3$.
(c) $\frac{p}{1+pq}$, which is $\frac{70}{121} \approx 0.579$ when $p = 0.7, q = 0.3$.
I'll work through part (b) to show you how the techniques in Concrete Mathematics work.
Suppose Player A chooses HTH and Player B chooses THH. Let $S_A$ be the sum of the winning configurations for Player A, so that
$$S_A = \text{HTH + HHTH + THTH + HHHTH + HTHTH + TTHTH} + \cdots$$
Similarly, the sum of the winning configurations for Player B is
$$S_B = \text{THH + TTHH + HTTHH + TTTHH} + \cdots$$
One advantage of doing this is that if we let $H = 0.7$ and $T = 0.3$ in these two equations $S_A$ and $S_B$ give the probabilities that Player A and Player B win, respectively.
Then, let $N$ denote the sum of the sequences in which neither player has won so far:
$$N = 1 + \text{H + T + HH + HT + TH + TT + HHH + HHT + HTT + THT + TTH + TTT} + \cdots$$
Now, we look for a set of equations relating $S_A, S_B,$ and $N$.
First, we can write the sum of all configurations in two different ways, so they must be equal:
$$1 + N(\text{H + T}) = N + S_A + S_B.$$
Adding HTH to any configuration in $N$ results in a win for $A$, a win for $A$ followed by TH, or a win for $B$ followed by a TH, so
$$N \text{ HTH} = S_A + S_A \text{ TH} + S_B \text{ TH}.$$
Finally, adding THH to a configuration in $N$ results in a win for $A$ followed by an H or a win for $B$,
so we have $$N \text{ THH} = S_A \text{ H} + S_B.$$
Letting $H = p$ and $T = q$ and solving the last three equations, I get $S_A = \frac{1-pq}{2-p}$ and $S_B = \frac{1-p+pq}{2-p}$. With $p = 0.7$, this yields $S_A = \frac{79}{130} \approx 0.608$ and $S_B = \frac{51}{130} \approx 0.392$.
For another example of the use of this technique, see a question related to two competing patterns in coin tossing.
First flip and second toss are independent events. So do first flip and second flip in the case that first flip is tail.
So use multiplication:
P(head on the first flip and 6 on the second tossing)=P(head on the first flip)*P(6 on the second tossing)=$\frac{1}{2}*\frac{1}{6}=\frac{1}{12}$
P(tails on both flip)=$\frac{1}{2}*\frac{1}{2}=\frac{1}{4}$
Win the game if either one of the two events happens, so use addition:
P(winning the game)=P(head on the first flip and 6 on the second tossing)+P(tails on both flip)=$\frac{1}{12}+\frac{1}{4}=\frac{1}{3}$
Best Answer
Flip the coin, obtaining some number of $\color{blue} T$'s until you see the first $\color{blue} H$. Once this happens, it becomes possible for Player A or B to win in the next two flips. Then flip the coin twice, obtaining a sequence of two letters. If we see $\color{red}{HH}$, Player A wins. If we see $\color{red}{TH}$, Player B wins. If we see $\color{red}{TT}$, then neither player can win in the next flip, so we start this process over. If we see $\color{red}{HT}$, then flip once more. Player B wins if we see $\color{green}{H}$; otherwise, we start over. For example, $$ \color{blue}{TTTTTTH} \color{red}{TH} $$ results in Player B winning, but $$ \color{blue}{TTTTTTTH} \color{red}{HT} \color{green}{T}$$ results in us starting over. The probability that Player A wins in one of these rounds is the probability of seeing $\color{red}{HH}$, which is $p^2$. The probability that Player B wins is the probability of seeing $\color{red}{TH}$ or $\color{red}{HT} \color{green}{H}$, which is $(1-p)p + p(1-p)p$. Therefore, if we repeat this until some player wins, the probability Player A wins is $$ \frac{p^2}{p^2 + (1-p)p + p(1-p)p} .$$