Calculate conditional expectation with respect to a sigma algebra

I am new to this stuff, so I have no idea how to really perform a calculation a conditional expectation of $E[X|F_1]$ and $E[X|F_2]$.

Let $\Omega=\{a,b,c\}$ , and $X$ is a random variable with $X(a) =1, X(b) =0 , X(c) =-1$.

$P(X=a)=P(X=b)=P(X=c)=1/3$ , so that $E[X] = 1*1/3+0*1/3+-1*1/3=0$

$F_1 \text{is the sigma algebra} = \{\text{null set} , \Omega \} . F_2=\{\text{null}, \Omega , \{a\},\{b,c\}\}.$

On my attempt , I think $E [X|F_1] (a)= E[X| Ω]= 0$ because $\{a\}$ only in $\{a,b,c\}$ in $F_1$, and $E[X|Ω]=EX=0$; similar for $E[X|F_1](b)$ and $E[X|F_1](c)$. I dont know this is correct logic or not.

for $E[X|F_2]$(a) , since a is in $\{a\}$ and $Ω$ in $F_2$

will $E[X|F_2](a)= X(a)*P({a})/P({a}) = 1$

I am really confused right now.

Also Can you tell me how to evaluate $E[E[X|F_1]|F_2]$ ?

For any discrete sigma-algebra $\mathcal G$, one can think of $E(X\mid\mathcal G)$ as being constant on every atom of $\mathcal G$ with the same expectation as $X$ on each of these atoms.

If $\mathcal G=\{\varnothing,\Omega\}$, there is only one atom, namely $\Omega$, hence $E(X\mid\mathcal G)$ is the constant random variable with value $E(X)$.

If $\mathcal G=\{\varnothing,A,\Omega\setminus A,\Omega\}$ where $0\lt P(A)\lt1$, the atoms are $A$ and $\Omega\setminus A$ hence $E(X\mid\mathcal G)$ is constant on $A$ and constant on $\Omega\setminus A$, that is, $E(X\mid\mathcal G)=x\mathbf 1_A+y\mathbf 1_{\Omega\setminus A}$ where $x=E(X\mid A)=E(X\mathbf 1_A)/P(A)$ and $y=E(X\mid\Omega\setminus A)=E(X\mathbf 1_{\Omega\setminus A})/P(\Omega\setminus A)$.

Finally, with the notations of the question, $E(X\mid\mathcal F_1)$ is constant, say $E(X\mid\mathcal F_1)=x$, hence $E(E(X\mid\mathcal F_1)\mid\mathcal F_2)=x$ for every $\mathcal F_2$, that is, $E(E(X\mid\mathcal F_1)\mid\mathcal F_2)=E(X)$.