I don't understand how to prove:
$(n-1)(n-2)(n-3)$ is $\Omega(n^3)$.
Also, what am I supposed to do here?
For each of the following predicates, determine, if possible, the smallest positive integer b where $n \ge b$ implies $P(n)$ appears to be true.
1) $log_2 n < n^{1/2}$
2) $n^{1/2} < n$
3) $F_n < 2^n$
4) $n! < n^n$
5) $2^n < n!$
6) $n < nlog_2n$
In order to prove that $f(x)$ is $\mathcal O (g(x))$ as $x\to \infty$, you need to prove that there exist two positive constants $c_1$ and $c_2$ and $x_0$:
$$\forall x>x_0\quad c_1 |g(x)|\le |f(x)|\le c_2 |g(x)|.$$
In the case of you polynomial $(n-1)(n-2)(n-3)$ it's evident that for $n\ge 3$ $$(n-1)(n-2)(n-3)\le n^3.$$ On the other hand, if $n\ge 4$ $$(n-1)(n-2)(n-3)=n^3 (1-1/n)(1-2/n)(1-3/n)\ge n^3 (1-1/4)(1-2/4)(1-3/4)=\frac{3}{32}n^3,$$ which allows to conclude that $(n-1)(n-2)(n-3)$ is $\mathcal O (n^3)$ as $n\to \infty$.
Apparently, the same reasoning allows to conclude that $n^3$ is $\mathcal O((n-1)(n-2)(n-3))$, so by Knuth's definition, you have also $(n-1)(n-2)(n-3)$ is $\Omega (n^3)$ as $n\to \infty$.
If you want to use Hardy's definition of $\Omega$, then you need to prove that $$\limsup_{n\to\infty} \frac{ (n-1)(n-2)(n-3) }{n^3}>0,$$ which is true because $\lim _{n\to\infty} \frac{ (n-1)(n-2)(n-3) }{n^3}=1$, so by this definition you have your $\Omega$ equivalence. I suggest you take a look at this wiki article.
As to you other questions, I think it's better to either make another post (one post - one topic of question), or at least provide us with your own results on the subject.