First part:
Let $a,b,c$ represent $3$ consecutive days. Since we are in state $1$, that means we have the sequence $(a,b) = \text{(no rain, rain)}$. In order to jump onto state $0$, there must hold $(b,c) = \text{(rain, rain)}$. Then we have the sequence $(a,b,c) = \text{(no rain, rain, rain)}$. According to the assumptions, starting from $(a,b)$ we can reach $c$ with probability $p=0.5$.
Also, $P_{11} = 0$. Why? If we still have $3$ consecutive days $a,b,c$ then it must hold $(a,b) = \text{(no rain, rain)}$ and $(b,c) = \text{(no rain, rain)}$, which can't happen.
Second part:
Notice that we start from state $0$, thus $\pi(0) = \begin{bmatrix} 1& 0 & 0 & 0\end{bmatrix}$ and we are going to evaluate the probability:
$$\pi(0)\cdot P^2 = \begin{bmatrix} 0.49 & 0.12 & 0.21 & 0.18\end{bmatrix}. $$
Thus, the probability that it rains on Thursday is going to be $p=0.49+ 0.12 = 0.61$ (see part $3$).
Third part:
From part $2$ it is known that the initial state is the state $1$. Assuming that we have the sequence $(a,b,c,d)$ with $a$ corresponding to the first day (Monday) and $d$ correspond to the last day (Thursday). Thus, we want the following to hold:
$$(a,b,c,d) = \text{(rain, rain, x, rain)}.$$ $x$ could either represent a rainy day or a non - rainy day. Thus, the are $2$ paths.
1: $(a,b,c,d) = \text{(rain, rain, rain, rain)}$
2: $(a,b,c,d) = \text{(rain, rain, no rain, rain)}$
Τhus, $(c,d)$ is going to be either (rain, rain), which indeed corresponds to state $0$ or (no rain, rain), which corresponds to state $1$.
Speaking with term of states the first $4-tuple$ corresponds to the path $0\to 0\to 0$, thus we have $p_{00}\cdot p_{00}= 0.7^2=0.49$ and the second $4-tuple$ corresponds to the path $0\to 2\to 1$, thus $p_{02}\cdot p_{21} = 0.3 \cdot 0.4 = 0.12$. Adding the two probabilities, leads us to the answer of the second part.
Best Answer
$$B\Rightarrow C\Rightarrow A\Rightarrow B$$ $$P(x_0=B,x_1=C,x_2=A,x_3=B)=P(x_0=B)\ P(x_1=C|x_0=B)$$ $$\qquad \qquad \qquad P(x_2=A|x_1=C)\ P(x_3=B|x_2=A)$$ $$P(x_0=B,x_1=C,x_2=A,x_3=B)=1\times 0.2\times 0.3\times 0.3=0.018$$ or $$B\Rightarrow A\Rightarrow C\Rightarrow B$$ $$P(x_0=B,x_1=A,x_2=C,x_3=B)=P(x_0=B)\ P(x_1=A|x_0=B)$$ $$\qquad \qquad \qquad P(x_2=C|x_1=A)\ P(x_3=B|x_2=C)$$ $$P(x_0=B,x_1=A,x_2=C,x_3=B)=1\times 0.3\times 0.5\times 0.3=0.045$$ and in total $$P=0.063$$ -----------Transition Matrix------------ \begin{array}{c|ccc} &A &B &C \\ \hline A &P(A|A) &P(B|A) &P(C|A)\\ B &P(A|B) &P(B|B) &P(C|B)\\ C &P(A|C) &P(B|C) &P(C|C) \end{array} which means if you select (2,1) it means that you look up for the probability of transition from B to A. Rows are for current state and columns for next state.