First, let's agree on notation: by $X_n$ we mean the state at "time" $n$; and we are told that the initial state ($n=0$) is $X_0 = v$. The transition matrix, denotes $M_{i,j} = P(X_{n+1}=j | X_{n}=i)$ (the row corresponds to the 'before' state, the column to the 'after' state).
For the first question, you want to compute a particular transition path. But remember that you start from $X_0$ (it can help to draw a graph of the transitions), so you actually are computing:
$$P(X_1=x,X_2=z,X_3=v | X_0=v)=\\
=P(X_1=x | X_0=v) P(X_2=z|X_1=x) P(X_3=v|X_2=z) =\\
= 0.6 \times 0.9 \times 0.7$$
(The first equation is true because it's a Markov chain).
For the second, you need to compute the probabilities of arriving of each one of the five states at time $n=4$. You could do that by summing all the paths that start from $X_0=v$, but that would be painful. (In general, perhaps not so much in this case because there are few transitions with positive probability). A more elegant way is to recall that the 4-step transition probabilities is given by $M^4$. Once you compute that, you just take the first row.
I believe that you can determine this by examining the eigenvalues of the transition matrix. A recurrent chain with period $d$ will have $d$ eigenvalues of magnitude $1$, equally spaced around the unit circle. I.e., it will have as eigenvalues $e^{2\pi ki/d} (0\le k<d)$.
The basic idea behind this is that if a recurrent Markov chain has period $d>1$, you can number the states so that its matrix has the block form $$\begin{bmatrix}
0 & P_1 & 0 & \cdots & 0 \\
0 & 0 & P_2 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & P_{d-1} \\
P_d & 0 & 0 & \cdots & 0
\end{bmatrix}.$$ If each of the $P_k$ is simply the scalar $1$, you have a simple permutation matrix with eigenvalues equal to the $d$th roots of unity. It’s fairly straightforward to prove that the block matrix above also has the $d$th roots of unity as eigenvalues.
When you have non-communicating sets of states as you do in the matrix $T$, you can analyze each set separately. The eigenvalues for the upper-left block are $1$ and $\frac{-5\pm i}{10}$, while the eigenvalues of the lower-right block are $1$ and $-\frac3{20}$. Neither of these blocks has roots of unity (other than $1$ itself) as eigenvalues, so the chain is aperiodic.
Best Answer
The formal way to do this and as defined in the book Introduction to Probability Models by Sheldon Ross is:
A state $i$ is recurrent if $\sum_{n=1}^{\infty}p_{ii}^n = \infty$
A state $i$ is transient if $\sum_{n=1}^{\infty}p_{ii}^n < \infty$
You can also define this as:
A state $i$ where $i \in S$ is said to be recurrent if $f_i$ = $P$(ever returns to $i$| starts in $i$) = 1
A state $j$ where $j \in S$ is said to be recurrent if $f_j$ = $P$(ever returns to $j$| starts in $j$) < 1
You would need to do this for all states in your matrix and solve it.