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.
For part $a)$,
Let $K_1$ denotes when coin $1$ is chosen and $K_2$ denotes when coin $2$ is chosen.
The probability that the coin lands on heads on exactly $7$ of the $10$ flips is $P(H)$
$$P(H)=P(H|K_1)P(K_1)+P(H|K_2)P(K_2)$$
$$P(H)=\dbinom{10}{7}(0.4)^7(0.6)^3(0.5)+\dbinom{10}{7}(0.7)^7(0.3)^3()0.5$$
For part $b)$,
When the first flip is head, then the probability is computed from choosing $6$ head flips out of $9$ flips in both coins case. So, we get $\dbinom{9}{6}(0.4)^6(0.6)^3(0.5)+\dbinom96(0.7)^6(0.3)^3(0.5)$
Let $T$ denotes the event that the first flip is head.
$$P(H|T)=\dfrac{P(H\cap T)}{P(T)}$$
$P(T)=P(T|K_1)P(K_1)+P(T|K_2)P(K_2)=(0.4)(0.5)+(0.7)(0.5)=0.55$
$P(H\cap T)=P(H\cap T|K_1)P(K_1)+P(H\cap T|K_2)P(K_2)$
$$=\dbinom96(0.4)^6(0.6)^3(0.5)+\dbinom96(0.7)^6(0.3)^3(0.5)$$
Best Answer
One of THHH or HHHH must occur first - it is impossible for neither to ever occur.
Complementary counting gives us
$$P\,\big(\text{THHH occurs before HHHH}\big) \,\,= \,\,1-P\,\big(\text{HHHH occurs before THHH}\big)$$
In order for HHHH to occur before THHH, the first four flips must all be H (think about why this must be the case). Given that, we can rewrite
$$ \begin{align*} P\,\big(\text{THHH occurs before HHHH}\big) \,\,&= \,\,1-P\,\big(\text{first four flips are all heads}\big)\\\\ &= \,\,1 - \left(\frac{1}{2}\right)^4\\\\ &= \,\,1 - \frac{1}{16}\\\\ &= \,\,\boxed{\,\frac{15}{16}\,\,} \end{align*} $$
If the coin is weighted with probability of heads $p\neq 1/2$, simply substitue $p$ for $1/2$ in the above calculation.