Expected number of comparisons in a merge step?
I will answer your first question about the expected number of comparisons from a statistics/probability standpoint because I think it's rather interesting and non-obvious. Let's assume the input is random, both arrays to be merged are the same length ($n/2$), and the merging algorithm is the generic step by step compare and insert method.

Let $C(i)$ be the number of ways to merge the two arrays of length $n/2$ in $n/2 + i$ comparisons.

Clearly $C(0) = 2$ because this means we are doing exactly $n/2$ comparisons then appending the rest of one list to the end of the other. If we have two lists $A = \{a_1, a_2, \ldots\}$ and $B = \{b_1, b_2, \ldots\}$, then these merges would look like:
$$[a_1, a_2, \ldots, a_{n/2}, b_1, b_2, \ldots, b_{n/2}]$$
or
$$[b_1, b_2, \ldots, b_{n/2}, a_1, a_2, \ldots, a_{n/2}]$$

We then have $C(1)$ is the number of ways we can nest $b_1$ within $[a_1, a_2, \ldots a_{n/2}]$ (and the mirror of this, nesting $a_1$ into $b_{1\ldots n/2}$). This can be expressed similarly to the number of integer solutions to equations. We can insert $b_1$ into $a_{1\ldots n/2}$ in the following locations represented by the symbol $\diamond$:
$$[\diamond a_1 \diamond a_2 \diamond \ldots \diamond a_{n/2 - 1} \diamond 
 a_{n/2}]$$
It is important to note that we can not place $b_1$ immediately after $a_{n/2}$ as that will create a redundant count to what we already counted in $C(0)$. So this reduces our number of locations from $n/2$ to $n/2-1$ which we can choose from.

Let $n/2 = k$ for simplicity. We can express $C(i)$ as a combinatorial expression:
$$C(i) = 2\binom{(k-1) + (i + 1) - 1}{(i+1) - 1} = 2\binom{k + i - 1}{i} = \frac{2(k+i - 1)!}{i!(k-1)!}$$

So we now have the total number ($N$) of possible merges is the total number of ways of choosing $n/2$ elements from a set of $n$. This can be seen easily if you imagine a skeleton array of length $n$, we then must select $n/2$ spots in this array to put one of our arrays being merged (and symmetrically the other array would be placed in the holes). We then have:
$$N = \binom{n}{n/2} = \binom{2k}{k}$$
This is again clear to see when we sum the $C(i)$s:
$$\begin{align}
\sum_{i=0}^{k-1} C(i) &amp; = \sum_{i = 0}^{k-1} 2 \binom{k + i - 1}{i}\\[0.5em]
&amp; = 2\binom{2k - 1}{k - 1}\\[0.5em]
&amp; = \frac{2(2k - 1)!}{k!(k-1)!}\\[0.5em]
&amp; = \frac{2k}{2k}\cdot \frac{2(2k - 1)!}{k!(k-1)!}\\[0.5em]
&amp; = \frac{(2k)!}{k!k!}\\[0.5em]
&amp; = \binom{2k}{k} = \binom{n}{n/2} = N
\end{align}$$

We then can find the probability of $k + i$ comparisons as:
$$\begin{align}
P[X = k + i] &amp; = \frac{C(i)}{N}\\[0.5em]
&amp; = 2\binom{k + i - 1}{i}\cdot \binom{2k}{k}^{-1}
\end{align}$$
Then it's easy to get the expected value:
$$\begin{align}
E[X] &amp; = \sum_{i = 0}^{k-1} (k + i) \cdot 2\binom{k + i - 1}{i}\cdot \binom{2k}{k}^{-1}\\[0.5em]
&amp; = \frac{2k^2}{k + 1}\\[0.5em]
&amp; = \frac{n^2}{n + 2} \approx n -2
\end{align}$$
From this it's clear to see that the expected number of comparisons is pretty close to $n - 2$. In fact we can pull up a table to see how they compare:
$$\begin{array}{|r|r|r|}
  \hline n &amp; \lfloor n^2/(n+2) \rceil &amp; n - 2\\ 
  \hline 2 &amp; 1 &amp; 0\\
  \hline 4 &amp; 3 &amp; 2\\
  \hline 8 &amp; 6 &amp; 6\\
  \hline 16 &amp; 14 &amp; 14\\
  \hline 32 &amp; 30 &amp; 30\\
  \hline 64 &amp; 62 &amp; 62\\
  \hline 128 &amp; 126 &amp; 126\\
  \hline 256 &amp; 254 &amp; 254\\
  \hline 512 &amp; 510 &amp; 510\\
  \hline 1024 &amp; 1022 &amp; 1022\\
  \hline
\end{array}$$
Another visual, here is the probability mass function of $x$ comparisons for merging into an array of size $1024$ (e.g. merging two arrays of size $512$):