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.
As stressed in the comments, the extra information available in part b) is that fact that the first toss gives you Heads. Conceptually, the problem is "how do you use the new information?". Informally, the fact that you see $H$ on the first trial increases the probability that you have coin #$2$. But by how much?
To answer that, suppose you toss each coin ten times. You'll see $4$ Heads from coin #$1$ and $7$ from coin #$2$. Thus you'll see $11$ Head all in all, $7$ of which come from coin #$2$. Thus, if all you are told was that you saw Heads, the probability that it came from coin #2 is $\frac 7{11}$.
Accordingly, you can now redo part a), looking for exactly $6$ Heads out of the next $9$ tosses where you have coin #$1$ with probability $\frac 4{11}$ and coin #$2$ with probability $\frac 7{11}$.
Best Answer
I wonder if the minuses in the 'official solution' should be equal signs.