F measurable and conditional expectation. 
(a):I found it easily cause sum of measurable sets are measurable. (b),(c): I know limsup(Sn/n) is also measurable but I can't prove that just sup(Sn/n) is measurable.
(d): I solved it by using the taking out what is known rule.
(e): I don't know whether I can take out X1 by taking out what is known rule or not. 
Many Thanks!

 A: (b) It is well-known that the countable supremum of measurable functions is measurable, i.e. if $(Y_n)_{n \in \mathbb{N}}$ is a sequence of $\mathcal{F}$-measurable (real-valued) random variables, then $\sup_{n \geq 1} Y_n$ is also $\mathcal{F}$-measurable. To see this, note that $$\left\{ \sup_{n \geq 1} Y_n > \alpha \right\} = \bigcup_{n \in \mathbb{N}} \{Y_n>\alpha\} \in \mathcal{F}$$ for any $\alpha \in \mathbb{R}$.

(e) We have
$$S_n = \mathbb{E}(S_n \mid \sigma(S_n)) = \sum_{j=1}^n \mathbb{E}(X_j \mid \sigma(S_n)).$$
If we can show that $$\mathbb{E}(X_j \mid \sigma(S_n)) = \mathbb{E}(X_1 \mid \sigma(S_n)) \qquad \text{for all $j=1,\ldots,n$}, \tag{1}$$ then we get
$$S_n = n \mathbb{E}(X_1 \mid \sigma(S_n)),$$
i.e.
$$\mathbb{E}(X_1 \mid \sigma(S_n)) = \frac{S_n}{n}.$$
It remains to prove $(1)$. To this end, recall that the $\sigma$-algebra $\sigma(S_n)$ is generated by sets of the form $\{S_n \in B\}$ for Borel sets $B$ and therefore $Y = \mathbb{E}(X \mid \sigma(S_n))$ if and only if $Y$ is $\sigma(S_n)$-measurable and
$$\int_{\{S_n \in B\}} X \, d\mathbb{P} = \int_{\{S_n \in B\}} Y \, d\mathbb{P}.$$
So, in order to show $(1)$, it suffices to show that
$$\int_{\{S_n \in B\}} X_1 \, d\mathbb{P} = \int_{\{S_n \in B\}} X_j \, d\mathbb{P}. \tag{2}$$
Denoting by $\mathbb{P}_{X_1,\ldots,X_n}$ the joint law of $(X_1,\ldots,X_n)$, we have
$$\int_{\{S_n \in B\}} X_1 \, d\mathbb{P} = \int 1_B(x_1+\ldots+x_n) x_1 \, d\mathbb{P}_{X_1,\ldots,X_n}(x_1,\ldots,x_n).$$
Now, since the random variables are independent and identically distributed, we get
$$\begin{align*} \int_{\{S_n \in B\}} X_1 \, d\mathbb{P} &= \int \dots \int 1_B(x_1+\ldots+x_n) x_1 \, d\mathbb{P}_{X_1}(x_1) \dots \, d\mathbb{P}_{X_n}(x_n) \\ &= \int \dots \int 1_B(x_1+\ldots+x_n) x_j \, d\mathbb{P}_{X_1}(x_1) \dots \, d\mathbb{P}_{X_n}(x_n) \\ &= \int 1_B(x_1+\ldots+x_n) x_j \, d\mathbb{P}_{X_1,\ldots,X_n}(x_1,\ldots,x_n) \\ &= \int_{\{S_n \in B\}} X_j \, d\mathbb{P} \end{align*}$$
for all $j=1,\ldots,n$.
A: To be honest, sometimes I feel measure theory or the concept measurable makes some simple probability more complex. 
Nevertheless, the problem are quite standard and I give the answer in the following. 
(b) Suppose $f_n(\omega)$ are all measurable function from $\mathcal{F}$ to $\mathbb{R}$, we aim to show $\sup_{n\geq m}f_n(\omega)$ are  measurable.  It is then enough to show  $\{\omega:\sup_{n\geq m}f_n(\omega)\leq q\}= \cap_{n\geq m}\{\omega:f_n(\omega)\leq q\}$. Since $f_n$ are measurable, we know $\{\omega:f_n(\omega)\leq q\}$ are measurable so is the intersection. Since we have $sup_{n\geq m}f_n(\omega)$ measurable now,$\{\omega:\sup_{n\geq m}f_n(\omega)> q\}$ is a measurable set. 
(e)Unfortunately, you can't. It is in the situation that you know at time $n$, the sum of the number of success $(X_i=1)$, and you try to guess the number of $X_1$, surely you can not determine $X_1$ in this case. 
Having understood the real situation, it is then reasonable to guess that $E(X_1|\sigma(S_n)=S_n/n$. You can then prove what @saz suggested or verify the definition of conditional expectation. 
$$ \int_{S_n=m} S_n/n dP =\int_{S_n=m} X_1 dP . $$
Left hand side is definitely $m/nP(S_n=m)=\frac{m}{n2^n}C^n_m,\forall m\leq n$ and $0$ otherwise. The right hand side can be computed by 
$$\int_{S_n=m} X_1 dP = \int 1_{S_n=m,X_1=1}dP =P(S_n=m,X_1=1)=\frac{1}{2^n}C_{m-1}^{n-1}. $$
A bit of simplification shows that the two are the same.
