Worst Case Scenario in MaxHeapify
1) This is the worst case in regards to all the possible recurrences and their respective sub-problem sizes. 

Let's take some small sample $n$'s for a tree of height $3$. This is to show how the varying worst-case sub-tree size ranges from 1/2 to 2/3.

$$
\begin{align}
n &amp;\quad \text{bottom half-full?} &amp; \text{size of worst case sub-tree} \\
8 &amp;\quad no &amp; 4/8 \\
9 &amp;\quad no &amp; 5/9 \\
10 &amp;\quad no &amp; 6/10 \\
11 &amp;\quad yes &amp; 7/11 \\
12 &amp;\quad no &amp; 7/12 \\
13 &amp;\quad no &amp; 7/13 \\
14 &amp;\quad no &amp; 7/14 \\
15 &amp;\quad no &amp; 7/15 \\
\end{align}
$$

We deal in worst case scenarios when defining running times. This is why we must look for the worst case sub-problem (sub-tree) to define the recurrence relation. Clearly with varying $n$ values, the recurrence would vary, so we must take the worst of them. This is clearly the result of the deepest level being half-full, resulting in the upper bound recurrence:

$$T(n) \leq T\left(\frac{2n}{3}\right) + \Theta(1)$$

Alternatively if we were looking for a lower bound, we should look for the best case of all those worst case sub problems. We could then achieve:

$$T(n) \geq T\left(\frac{n}{2}\right) + \Theta(1)$$

2) The heap with the bottom level half full will always have the worse running time no matter how trivial the difference is. Minor differences like the one you described are not important. We're not comparing for one instance, time complexity is used to compare on a larger scale. On a large scale you can see the size of the sub problem ranges from n/2 to 2n/3.