It is known that $X^2(\omega)$ is a random variable on $(\Omega, \: \mathcal{F}, \: \mathbb{R} )$
Are $X(\omega)$ and $|X(\omega)|$ also random variables? I knew that if $X(\omega)$ is a random variable then $X^2(\omega)$ also is one. I've looked up the definition but I'm still confuse.
In general it does not hold that if $X^2$ is random variable then $X$ is a random variable.
If $\mathcal F=2^\Omega$, then all functions $F:\Omega\rightarrow\mathbb R$ are random variables, being measurable. In particular, $X$ is random variable.
If $\mathcal F\ne2^\Omega$ we can construct an example which does not hold as following:
Let $A\subseteq\Omega$ non-measurable ($A\not\in\mathcal F)$. Let $X:\Omega\rightarrow\mathbb R$ defined by:
$$X(\omega)= \begin{cases} 1&\text{ if }x\in A\\ -1&\text{ if }x\not\in A\\ \end{cases} $$
Then $X$ is not measurable ($A=X^{-1}(\{1\})\not\in\mathcal F)$, so $X$ is not a random variable, but $X^2=1$ is a random variable.
For your second question, $|X|$ can be written as the composition of the two measurable functions $X^2:\Omega\rightarrow\mathbb R$ and $f:\mathbb R\rightarrow\mathbb R$, $f(x)=\sqrt x$ (the last function is measurable because it is continuous). Therefore, if $X^2$ is a random variable, then $|X|$ is a random variable.