The transition matrix $P$ has entries $p_{ij}$ which are the probability of transitioning from $i$ to $j$. So if your states are listed as "sunny,partly cloudy,rainy" then
$$P=\begin{bmatrix} 0.5 & 0.4 & 0.1 \\
0.4 & 0.5 & 0.1 \\
0.2 & 0.2 & 0.6 \end{bmatrix}$$
Given an initial distribution $\lambda$ (written as a row vector), the distribution at time $n$ is $\lambda P^n$. (If you prefer, you can work with a column vector and deal with $(P^T)^n \lambda$.) So you can find your second answer by taking $\lambda=[0,0,1]$ and calculating $\lambda P^2$.
The first answer is slightly different because you are being asked about the probability that it rains two days in a row. Here you'll want to use the independence: the probability of being rainy for the next two days is the probability to go from rainy to rainy, and then rainy to rainy again.
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
Denoting your state space by $\{S,C,R\}$, you are searching for:
$$\Bbb P (X_4 = R, X_3 = R | X_1 = S) = \Bbb P(X_4 = R | X_3 = R,X_1=S)\Bbb P (X_3=R|X_1=S)$$ Can you go further by yourself? This step should be an answer to your question already.
EDIT: Further, by the markov property above equals $$ \Bbb P(X_4 = R | X_3 = R)\Bbb P (X_3=R|X_1=S) = P(R,R)\cdot P^2(S,R) = 0.1\cdot 0.21 $$