Typical implementations of quicksort don't append the pivot to one of the lists. Instead, they are implemented (in Haskell notation) as
qsort [] = []
qsort xs = qsort left ++ [pivot] ++ qsort right
where pivot = head xs
left = filter (<x) xs
right = filter (>=x) xs
which keeps the symmetry between the left and right lists (if the list elements are distinct) and results in the same number of comparisons for an initial list that is sorted increasing or sorted decreasing.
Note that the 'worst case' isn't just found for the lists that are sorted in increasing or decreasing order. For example, the following lists all require six comparisons to sort:
[1,2,3,4]
[1,2,4,3]
[1,4,2,3]
[1,4,3,2]
[4,3,2,1]
[4,3,1,2]
[4,1,2,3]
[4,1,3,2]
In general, any ordering which results in the lest or greatest element being selected as the pivot at every stage of the recursion will give you worst case performance.
The bound $m+n-1$ is achieved by a linear merge: Start with a pointer at the beginning of each list, compare the two elements pointed to, add the smaller one to the merged list and increase that pointer. In case you have to compare the last two elements, you've made $(m-1)+(n-1)$ comparisons to get to that point, so the worst case takes $m+n-1$ operations.
[Edit in response to the edit in the question:]
For simplicity, let's assume that all values are different so that all comparisons have a binary result. (Since this is about worst-case behaviour, we can make arbitrary assumptions about the input.) With $k$ binary results, you can distinguish at most $2^k$ different cases. The result of merging two sorted lists of lengths $m$ and $n$ can be specified by the positions of the $m$ elements of the one list in the merged list, and there are $\binom{m+n}m$ different arrangements of these positions. Thus you need at least $\left\lceil\log_2\binom{m+n}m\right\rceil$ comparisons. That solves cases c) and d), since
$$\left\lceil\log_2\binom{3+4}3\right\rceil=\left\lceil\log_235\right\rceil=6=3+4-1$$
and
$$\left\lceil\log_2\binom{4+4}4\right\rceil=\left\lceil\log_270\right\rceil=7=4+4-1\;.$$
Case b) needs a bit more work, since
$$\left\lceil\log_2\binom{2+4}2\right\rceil=\left\lceil\log_215\right\rceil=4\lt2+4-1\;.$$
To allow the merge to be done with $4$ comparisons, the first comparison would have to divide the $15$ possibilities into two groups of $8$ and $7$; otherwise one of the groups would have more than $8$ possibilities and couldn't be resolved with the remaining $3$ comparisons. Due to the symmetry between greater and lesser, we only have to consider comparisons involving the first element of the two-element list as first comparisons. Comparing the first element of the two-element list with one of the four elements of the other list divides the $15$ possibilities into groups of $10/5$, $6/9$, $3/12$ and $1/14$, respectively. Since there's a group with more than $8$ elements in each case, it's impossible to do the merge with $4$ comparisons in all cases.
Best Answer
It looks like you have a mistake in (1). You wrote: $$1 + 1 + \cdots + 1 = \sum_{i=1}^{n-1} i = (n-1).$$
That sum should have been: $$\sum_{i=1}^{n-1} 1 = n-1.$$
Now in terms of the comparisons, those get made when percolating the element $$A[i]$$ forward. If the elements are already in ascending order, then $$A[i] \geq A[i-1]$$ and no swap is made. So the inner loop does not execute.