Consider the ways to get heads on both of the first two tosses: you can have the unfair coin, or you can have a fair coin and toss heads twice in a row. The probability of having the unfair coin is $\frac13$. The probability of having a fair coin and tossing heads twice in a row is $\frac23\cdot\frac14=\frac16$. If you know that you’ve tossed heads twice in a row, then you know that you’re in one of these two cases; the probability that you’re in the first case (have the unfair coin) is $$\frac{\frac13}{\frac13+\frac16}=\frac23\;,$$ and the probability that you’re in the second case (have a fair coin) is $1-\frac23=\frac13$.
You know the probability of getting tails with the unfair coin and the probability of getting tails with a fair coin, so you can now calculate the overall probability of getting tails if you’ve tossed heads on your first two tries; just use the same basic approach that you used in the first three questions.
If you began by tossing tails, twice, on the other hand, you know that you have a fair coin.
Let $A_{m,n}$ be the probability that player $A$ flips two consecutive heads before player $B$ flips three consecutive heads, given that $A$'s (resp. $B$'s) current run of heads has length $m$ (resp. $n$). Then
$$
A_{m,n}=\frac{1}{4}\left(A_{m+1,n+1} + A_{m+1,0} + A_{0,n+1}+A_{0,0}\right);
$$
the boundary conditions are that $A_{m,n}=1$ for $m\ge 2$ and $n < 3$, and $A_{m,n}=0$ for $m \le 2$ and $n\ge 3$. You want to find $A_{0,0}$. The relevant six equations are:
$$
\begin{eqnarray}
A_{0,0} &=& \frac{1}{4}A_{1,1} + \frac{1}{4}A_{1,0} + \frac{1}{4}A_{0,1} + \frac{1}{4}A_{0,0}\\
A_{0,1} &=& \frac{1}{4}A_{1,2} + \frac{1}{4}A_{1,0} + \frac{1}{4}A_{0,2} + \frac{1}{4}A_{0,0} \\
A_{1,0} &=& \frac{1}{2} + \frac{1}{4}A_{0,1} + \frac{1}{4}A_{0,0} \\
A_{1,1} &=& \frac{1}{2} + \frac{1}{4}A_{0,2} + \frac{1}{4}A_{0,0} \\
A_{0,2} &=& \frac{1}{4}A_{1,0} + \frac{1}{4}A_{0,0} \\
A_{1,2} &=& \frac{1}{4} + \frac{1}{4}A_{0,0},
\end{eqnarray}
$$
or
$$
\left(\begin{matrix}3/4 & -1/4 & -1/4 & -1/4 & 0 & 0 \\
-1/4 & 1 & -1/4 & 0 & -1/4 & -1/4 \\
-1/4 & -1/4 & 1 & 0 & 0 & 0 \\
-1/4 & 0 & 0 & 1 & -1/4 & 0 \\
-1/4 & 0 & -1/4 & 0 & 1 & 0 \\
-1/4 & 0 & 0 & 0 & 0 & 1
\end{matrix}\right)\times\left(\begin{matrix} A_{0,0} \\ A_{0,1} \\ A_{1,0} \\ A_{1,1} \\ A_{0,2} \\ A_{1,2}
\end{matrix}\right)
= \left(\begin{matrix}
0 \\
0 \\
1/2 \\
1/2 \\
0 \\
1/4
\end{matrix}\right),
$$
assuming no typos. Further assuming no typos entering this into WolframAlpha, the result is
$$
A_{0,0} = \frac{1257}{1699} \approx 0.7398,
$$
which at least looks reasonable.
Update: As pointed out in a comment, the above calculation finds the probability that $A$ gets two consecutive heads sooner than $B$ gets three consecutive heads; the original problem asks for the opposite. The correct boundary conditions for the original problem are that $A_{m,n}=1$ for $m<2$ and $n\ge 3$ and $A_{m,n}=0$ for $m\ge 2$ and $n\le 3$. The matrix equation becomes
$$
\left(\begin{matrix}3/4 & -1/4 & -1/4 & -1/4 & 0 & 0 \\
-1/4 & 1 & -1/4 & 0 & -1/4 & -1/4 \\
-1/4 & -1/4 & 1 & 0 & 0 & 0 \\
-1/4 & 0 & 0 & 1 & -1/4 & 0 \\
-1/4 & 0 & -1/4 & 0 & 1 & 0 \\
-1/4 & 0 & 0 & 0 & 0 & 1
\end{matrix}\right)\times\left(\begin{matrix} A_{0,0} \\ A_{0,1} \\ A_{1,0} \\ A_{1,1} \\ A_{0,2} \\ A_{1,2}
\end{matrix}\right)
= \left(\begin{matrix}
0 \\
0 \\
0 \\
0 \\
1/2 \\
1/4
\end{matrix}\right),
$$
The result becomes $A_{0,0}=\frac{361}{1699}\approx 0.2125$. The two results add to slightly less than one because there is a nonzero probability that both players hit their goals at the same time... this probability is $81/1699\approx 0.0477$.
Best Answer
the answer is indeed $\frac 12$ .
As an alternative way to see that: let's pause just before Bob tosses his final (extra) toss. At this point, there are three possible states: either Bob is ahead, Alice is ahead, or they are tied. Let $p$ be the probability that Bob is ahead. By symmetry, $p$ is also the probability that Alice is ahead (so the probability of a tie is $1-2p$). Note that symmetry does clearly apply here since they have thrown the same number of tosses. Bob has exactly two ways to win: either he is ahead before the last toss, or they are tied and Bob gets $H$ on the last throw. Thus the probability that Bob eventually wins is $$p+\frac 12 \times (1-2p)=p+\frac 12 -p =\frac 12$$