We toss a fair coin three times. So the probability space is given by $(\Omega ^3, \mathcal F^{\otimes 3},\mathbb P^{\otimes 3})$ where $\Omega =\{H,T\}$, $\mathcal F=2^\Omega $ and $\mathbb P\{\omega \}=1/2$.
Let the $\sigma -$algebra $$\mathcal F_1=\sigma \{A_1,A_2\},$$ where $A_1=\{(H,T,T),(H,T,H),(H,H,T),(H,H,H)\}$ and $A_2=\{(T,T,T),(T,T,H),(T,H,T),(T,H,H)\}$,
Let $X:\Omega ^3\to \mathbb R^3$ defined by $X=(X_1,X_2,X_3)$ where $X_i:\Omega \to \mathbb R$ are defined by $X_i=\boldsymbol 1_{\{H\}}$.
I know $$\mathbb E[X\mid \mathcal F_1](\omega _1,\omega _2,\omega _3)=\begin{cases} \frac{1}{\mathbb P(A_1)}\int_{A_1}Xd\mathbb P^{\otimes 3}&\omega _1=H\\ \frac{1}{\mathbb P(A_2)}\int_{A_2}Xd\mathbb P^{\otimes 3}&\omega _1=T, \end{cases}$$ but I don't know how to compute those quantity. I think $\mathbb P(A_i)=\frac{1}{2}$, but I'm not sure how to compute $\int_{A_1}X d\mathbb P^{\otimes 3}$. Is it just $$\mathbb P\{X_1=1, X_2=0, X_3=0\}+\mathbb P\{X_1=1,X_2=1,X_3=0\}+\mathbb P\{X_1=1, X_2=0,X_3=1\}+\mathbb P\{X_1=1,X_2=1,X_3=1\} \ \ ?$$ If yes, by independence of $X_i$ I find $4\cdot \frac{1}{8}$ and thus $$\mathbb E[X\mid \mathcal F_1]=2\cdot \frac{1}{2}=1$$ that look quite strange for me. It looks quite complicated for such an "easy" problem.