Does this Markov chain have a unique stationary distribution? Or are there more than one? I know that a positive recurrent Markov chain has a unique distribution but I am not sure how to check if this Markov chain is positive recurrent.
The chain is irreducible on a finite state space hence it is positive recurrent and has a unique stationary distribution $\pi$.
Following the precise indications in the comments, one gets that $\pi$ is in the proportions 3:2:1, thus $\pi=(\frac12,\frac13,\frac16)$.
The one-step analysis at the end of the question is the wrong way around. You need a fixed target, and the index on $E$ should index the states from which you're trying to reach that target.
But in the present case there are less cumbersome ways to get the expectation values you want.
For the first question, the stationary distribution is constant by symmetry, so $\mu_1=4$ is immediate. (Note that this counts staying at $1$ as a "return".)
For the second question, get rid of the self-loops and scale the number of steps by $\frac1{1-\frac14}=\frac43$ to compensate. Then you have a simple walk on the given graph, with equal probabilities $\frac12$ to go either way. Combine the steps into pairs. Each pair has probability $\frac12$ to return to $1$ and $\frac12$ to get you to $4$. Thus the expected number of pairs is the expected number of trials until the first success in a Bernoulli trial with success probability $\frac12$. So the expected number of steps from $1$ to $4$ is
$$\frac43\cdot2\cdot\frac1{\frac12}=\frac{16}3\;.$$
Here's the one-step analysis:
Let $E_n$ be the expected number of steps from state $n$ to state $4$. We're looking for $E_1$. The $E_i$ satisfy
\begin{align}
E_1&=1+\frac14E_1+\frac38E_2+\frac38E_3\;,\\
E_2&=1+\frac14E_2+\frac38E_1+\frac38E_4\;,\\
E_3&=1+\frac14E_3+\frac38E_1+\frac38E_3\;,\\
E_4&=0\;.
\end{align}
(I wrote it out in full since you wanted the general method; in the present case, we could use the fact that $E_2=E_3$ by symmetry to save a variable.) Solving this system of equations yields $E_1=\frac{16}3$, as derived above.
Best Answer
You can replace any of the four equations (e.g. last one) with the last equation $\pi_1+\cdots+ \pi_4=1$ to form the augmenterd matrix (note all $\pi$s were transfered to the left): $$ \left[ \begin{array}{cccc|c} -0.3&0&0&0.03&0\\ 0.06&-0.3&0.15&0.03&0\\ 0.18&0.18&-0.3&0.24&0\\ 1&1&1&1&1\\ \end{array} \right] $$ Can you solve it using the Gauss elimination method?