Determining the stable (long term) state probabilities
The solutions of the following system of equations are the stationary state probabilities:
$$\begin{matrix}[\pi_A\ \pi_B \ \pi_C]\\
\\
\\
\end{matrix}
\begin{bmatrix}
0.2 & 0.4 & 0.4 \\
0.4 & 0.3 & 0.3 \\
0.6 & 0.2 & 0.2 \\
\end{bmatrix}
\begin{matrix}=[\pi_A\ \pi_B \ \pi_C]\\
\\
\\\end{matrix}
$$
and
$$ \pi_A+\pi_B+\pi_C=1 .$$
Note
Here $\pi_A\not=1$. $\pi_A$ is a hypothetical quantity an intuitive definition of which (together with that of $\pi_B$,$\pi_C$) is as follows: "If we started the system choosing a state randomly with probabilities $\pi_A,\pi_B, \pi_C$ then the probability that the next state would be $A$, $B$, or $C$ would be $\pi_A,\ \pi_B, \pi_C$, respectively." Such stationary probabilities do not necessarily exists. You can read about the details here.
Continued
Having performed the vector-matrix multiplication we arrive at a redundant system of equations. After omitting the redundant equation and after some rearrangement we have
$$\begin{matrix}
-0.8\pi_A & +& 0.4\pi_B & +& 0.6\pi_C &=0 \\
\ \ 0.4\pi_A & -& 0.7\pi_B& +& 0.2\pi_C&=0 \\
\pi_A&+&\pi_B&+& \pi_C&=1&
\end{matrix}$$
The solution, with the help of Alpha, is $$\pi_A=\frac{5}{13}, \ \pi_B=\frac{4}{13},\ \pi_C=\frac{4}{13} .$$
Question a.
We can toss heads in each states, so
$$P(\text{heads})=P(\text{heads}|\text{in state} A)\pi_A+P(\text{heads}|\text{in state}B)\pi_B+P(\text{heads}|\text{in state}C)\pi_C=\frac{2}{10}\frac{5}{13}+\frac{4}{10}\frac{4}{13}+\frac{6}{10}\frac{4}{13}=\frac{5}{13}=\pi_A.$$
That is, the result of the tossing is a head in $\frac{5}{13}100$% of the time on the long run.
Remark
It seemed to be accidental that
$$P(\text{heads})=\pi_A.$$
I wondered, if it was? So I took $p_A, p_B, \text{ and }p_C$ as the probabilities of the heads in the states $A, B, \text{ and } C$, respectively.
In order to get the stationary probabilities we have to solve the following system of equations:
$$\begin{matrix}
p_A\pi_A & +& p_B\pi_B & +& p_C\pi_C &=\pi_A \\
\frac{1-p_A}{2}\pi_A & +& \frac{1-p_B}{2}\pi_B& +& \frac{1-p_C}{2}\pi_C&=\pi_B \\
\pi_A&+&\pi_B&+& \pi_C&=1&
\end{matrix}.$$
The solutions are
$$
\begin{matrix}
\pi_A=\frac{p_B+p_C}{-2p_A+p_B+p_C+2}\\
\pi_B=\frac{1-p_A}{-2p_A+p_B+p_C+2}\\
\pi_C=\frac{1-p_A}{-2p_A+p_B+p_C+2}
\end{matrix}.
$$
At this point, note that $p_A<1$. If $p_A$ was $1$ then the Markov chain would never get out of state $A$. Then, of course,
$$1=P(\text{heads})=\pi_A.$$
If $p_A<1$ then the solutions above are valid and we may compute the probability of the heads in general. Watch this:
$$P(\text{heads})=\frac{(p_B+p_C)p_A+(1-p_A)p_B+(1-p_A)p_C}{-2p_A+p_B+p_C+2}=\pi_A \ !$$
Perhaps this is what the OOOOOP wanted to see.
Question b.
We have to go back to the initial state which is $A$. Now, the following vector-matrix product will give the probabilities that we end up in states $A$,$B$, or $C$ at the $\text{n}^{th}$ minute:
$$\begin{matrix}[1\ 0 \ 0]\\
\\
\\\end{matrix}
\begin{bmatrix}
0.2 & 0.4 & 0.4 \\
0.4 & 0.3 & 0.3 \\
0.6 & 0.2 & 0.2 \\
\end{bmatrix}^n\begin{matrix}=[\pi_A^n\ \pi_B^n \ \pi_C^n]\\
\\
\\\end{matrix}
$$
Finally, we have to repeat the steps performed during solution a. But this time we work with the results of the vector-matrix product above with $n=3$ and with $n=10$.
Let $p$ be the probability you want.
Look at the possible sequences of three tosses (including the initial $H$). We have $\{HH*,HTT,HTH\}$ Conditional on the first being $H$, the probability of $HH*$ is $\frac 12$, and the probability of the other two are each $\frac 14$. The first is a win for $H$, the second is a win for $T$, and the third restarts the game. Thus $$p=\frac 12\times 0+\frac 14 \times 1 +\frac 14\times p\implies p=\frac 13$$
Note: if you want to allow for a weighted coin, the same calculation works but we need to correct for those initial probabilities. Thus assume that the coin comes up $H$ with probability $\phi$. Then $HH*$ has probability $\phi$, $HTT$ has probability $(1-\phi)^2$, and $HTH$ has probability $(1-\phi)\phi$. So the recursion is now $$p=\phi \times 0+ (1-\phi)^2 \times 1 + (1-\phi)\phi \times p\implies p=\frac {(1-\phi)^2}{1-(1-\phi)\phi}$$ Comparison shows that this matches the result obtained by looking at infinite sums, as in the posted solution of @BCLC (with different notation, unfortunately).
Best Answer
The transition matrix reads: $P(X_n=0\to X_{n+1}=0)=\frac12$, and $P(X_n=0\to X_{n+1}=1)=\frac12$, and $P(X_n=0\to X_{n+1}=2)=0$. This is me reading off the first row.
In the context, this means the probability of going from $0$ successive heads to $0$ successive heads is $\frac12$ which makes sense, since this is the probability of rolling a tails. The probability of going from $0$ successive heads to $1$ successive head is also $\frac12$ since this is probability of hitting heads. Finally, probability of going from $0$ successive heads to $2$ successive heads is obviously not possible in one toss, so this probability is $0$.
Similarly, you can read off the other two rows.
Judging by the last row, this game ends when you get $2$ successive heads, since from here it can never go back down to $0$. I suppose this must have been in the question.
EDIT:
In an attempt to explain this better, I will also go through the second row.
Firstly, we must recognise what the columns and rows mean. The row represents the state in which we start - so the first row represents starting in the state of having $0$ successive heads, etc. The column represents the state in which we finish after $1$ toss of a coin. So the second column represents finishing with $1$ successive head, which means you used to have none, and you have just rolled a heads.
Now, the second row is $$\left(\begin{matrix}\frac12 & 0 & \frac12 \end{matrix}\right)$$ The first element in this represents the probability of going from the state of $1$ successive heads to $0$ successive heads (row $2$ corresponds to starting with $1$ successive heads, column $1$ corresponds to ending with $0$ successive heads). Clearly, to go from having $1$ successive heads to having none means we rolled a tails, which has probability $\frac12$.
The second element in this represents the probability of going from the state of $1$ successive heads to staying in this state. But we must toss the coin, and as I wrote in the comments, no matter what we get from the toss, we must change - rolling heads means we now have $2$ successive heads, rolling tails means we now have none (as described above). So we see it is impossible to stay in this state, hence why this entry is $0$.
Finally the third element represents the probability of going from $1$ successive heads to $2$. This of course happens when we roll a heads, which has probability half.
I hope this helped to simplify it.