I am trying to study the asymptotics of $f(n) = {n \choose n/2}$, $g(n) = n!$ and $h(n) = {n \choose 3 }$.
First, $h(n)$ is easy as $h(n) = O(n^3)$ and we see that since $n! \approx \sqrt{2 \pi n} (n/e)^n$, then
$$ g(n)/h(n) \approx n^{-2.5} (n/e)^n = \dfrac{ n^{n-2.5} }{e^n } $$
HOwever, I dont see a way to prove $g/h \to \infty$ to prove what is intuitively clear: $n! >> n^3$. What I am having difficulties is with $f(n)$. My thought is to write
$$ f(n) = \dfrac{ n! }{(n/2)! (n/2)! } = \dfrac{ n(n-1)(n-2).... 1 }{(n/2)^2 (n/2-1)^2 ...1} $$
And I see that $f(n) $ is approximately $0$ because powers in the denominator are getting higher. However, this is wrong, since for instance ${1000 \choose 500} $ is a very large number. What is happening here?
For your first question, note that $$\frac{g(n)}{h(n)} = \frac{6n!}{n(n-1)(n-2)} = 6(n-3)!$$ No need to use Stirling's approximation.
For your second question your misunderstanding comes from not actually counting how many factors are in the numerator and denominator. Although the denominator does have higher powers of $n/2$ and $n/2-1$ and so on, the numerator still has larger terms like $n$ and $n-1$. And both the numerator and denominator has $n$ factors (if you write $(n/2)^2$ as $(n/2)(n/2)$). There are a lot of things going on here, so it is difficult to discern the behavior just by looking at that expression.
Stirling's approximation can help a lot here. Since $n! \approx \sqrt{2 \pi n}(n/e)^n$ and $(n/2)! \approx \sqrt{2\pi n/2} (n/(2e))^{n/2}$ we have $$\binom{n}{n/2} = \frac{n!}{(n/2)!(n/2)!} \approx \frac{\sqrt{2}}{\sqrt{\pi n}}2^n$$