a. Consider the same problem but we require the numbers at the end to be in ascending order (i.e. descending order is not allowed). Let $m_k$ be the least number of moves needed to guarantee that this can be done when there are $k$ numbers. By symmetry, this number is the same if we require the numbers at the end to be in descending order. Clearly, we have $n_k \le m_k$.

We claim that $m_{k+1} \le m_k + 1$ for $k \le 4$. Given a permutation of $1, 2, \dots, k+1$, one of the numbers $1$ and $k+1$ must be located at the first two or the last two positions since there are at most $5$ numbers in total. We can use at most $1$ move to put this number to the first or the last position. For the remaining numbers, we need at most $m_k$ moves to sort them in the correct order, so that all numbers are in monotonic order. This proves the inequality.

It follows that $n_4 \le m_4 \le m_3 + 1 \le m_2 + 2 = 3$ (obviously, $m_2 = 1$). Next, we show that $2413$ cannot be changed to $1234$ or $4321$ in $2$ steps. This can be verified by checking all possibilities as follows. (Note that swapping the same block twice is useless.)

$$\begin{array}{ccc} \underline{2413} \longrightarrow 4213 \longrightarrow 4123, 4231, 1243 \\ 2413 \longrightarrow 2143 \longrightarrow 1243, 2134 \\ \underline{2413} \longrightarrow 2431 \longrightarrow 4231, 2341, 2134 \end{array}$$

Therefore, we must have $n_4 = 3$.

b. We have $n_5 = 4$.

Again, we list out all possibilities.