First a little arrangement of the states let you to have a better picture for the classification

\( \begin{matrix} 1 \\ 3 \\ 2 \\ 4 \end{matrix}

\begin{bmatrix} 0.7 && 0.3 && 0 && 0 \\

0.3 && 0.7 && 0 && 0 \\

0.4 && 0.4 && 0.2 && 0 \\

0.4 && 0.4 && 0 && 0.2 \end{bmatrix}\)

So it is easily see that states 1, 3 are not communicate with states 2, 4 and thus this markov chain is reducible.

A simple check for the aperiodic state is that you see the transition probabilities in the diagonal are non-zero, i.e. it is always possible to return to the same state in the next step.

The reasoning is completely correct. So if the question only ask you which state without proof, maybe this is enough for you to write them down immediately.

http://en.wikipedia.org/wiki/Markov_chain#Recurrence
A more formal argument is to check the definition. E.g. for state 2, because it is not accessible from other states as you have mentioned, so the first return time is either 1 or infinity (never return). So the sum = 0.2 < 1 which shows that it is transient.

For the class 1, 3, you can show that the \( f_{11}^{(n)}, f_{33}^{(n)} \) are just the geometric distribution pmf and of course the sum equals to 1, which means they are recurrent.