The solution for $(c)$ is perfect up to $\mu^{*}(C_n\setminus E_n) = \mu^{*}(C_n) - \mu^{*}(E_n) \leq \epsilon/2^{n}$, so let's keep going from here.
It is easy to lose track of what we are doing with these kind of problems, so let's not forget that we are trying to approximate the set $E$ from outside, and that right now we only have an approximation for every $n$. The obvious thing to do is then to consider the union of the approximating sets: let $$C_{\epsilon} = \bigcup_{n = 1}^{\infty}C_n.$$
The $\epsilon$ in $C_{\epsilon}$ is there to stress the dependence upon $\epsilon$ of the $C_n$'s and hence of $C$. It is worth noticing that $C_{\epsilon} \in \mathcal{A}_{\sigma}$ being the countable union of countable unions of elements of $\mathcal{A}$.
For the next step to work you might want to state clearly in your proof that we can take $C_n \subset \cup_{i = 1}^nX_i$.
Intuitively, $C_{\epsilon}$ is a decent approximation of $E$. Let's make this precise:
\begin{align}
\mu^*(C_{\epsilon} \cap E^c) = & \mu^*\Big(\bigcup_{n = 1}^{\infty} (C_n \cap E^c)\Big) = \mu^*\Big(\bigcup_{n = 1}^{\infty} (C_n \cap E_n^c)\Big) \\ \le & \sum_{n = 1}^{\infty}\mu^*(C_n \setminus E_n) \le \epsilon.
\end{align}
(good job considering the $2^{-n}$, that really came in handy!)
Let's go back to our intuition: $C_{\epsilon}$ is an $\epsilon$-good approximation of $E$, therefore we would like to send $\epsilon$ to $0$ to get the best possible approximation. Since we want to end up in $\mathcal{A}_{\sigma \delta}$ what we need to do is to "discretize" the sets $C_{\epsilon}$ considering instead the sets $C_{\frac{1}{n}}$. (the notation is unfortunate, these resemble too much the $C_n$ sets, hopefully this won't cause any confusion)
Then clearly $$C := \bigcap_{n = 1}^{\infty}C_{\frac{1}{n}} \in \mathcal{A}_{\sigma \delta}$$ and satisfies $\mu^*(C \setminus E) = 0$, proving the claim.
To show this last equality notice that $$\mu^*(C \setminus E) \le \mu^*(C_{\frac{1}{n}} \setminus E) \le \frac{1}{n} \to 0.$$
Let me know if there is anything that I need to clarify!
Note that one must be careful about what ae. means. When we say that $f=g$ ae., it means that there is a measurable set $E$ of measure zero such that
$f(x)=g(x)$ for $x \notin E$.
Suppose $\mu$ is complete. Let $E$ be the exceptional set where $f(x) \ne g(x)$.
Suppose $A$ is measurable. Then $g^{-1}(A) = ( g^{-1}(A) \cap E) \cup ( g^{-1}(A) \cap E^c)$. The set $g^{-1}(A) \cap E$ is measurable since
it is contained in $E$ which has measure zero.
We have $g^{-1}(A) \cap E^c= f^{-1}(A) \cap E^c$, hence it is measurable
and so $g^{-1}(A)$ is measurable, and so $g$ is measurable and so Part (a)
holds.
Now suppose Part (a) holds. Let $N \subset E$, where $E$ has measure zero. Let
$f=1_{E}$ and $g = 1_{N}$. Then $f=g$ ae. and so $g$ is measurable. Hence
$g^{-1}(\{1\}) = N$ is measurable. Hence $\mu$ is complete.
Part (b) is similar. Note that if $h_n \to h$ with the $h_n$ measurable, then
$h$ is measurable. So, this really can be reduced to Part (a) fairly easily.
Best Answer
Uhm, I only have a suggestion for part a. Notice that $E_{\epsilon,\delta}$ almost look like the inverse image of an open set. I know, $\leq$ does not make it look like the inverse image of an open set. But if you divide the set in $E^1$ and $E^2$, where for $E^1$ the strict inequality holds, and for $E^2$ equality holds, then $E^1\cap E^2=\emptyset$, and $E^1$ IS the inverse image of an open set. As for $E^2$, well, can there actually be any element in there?
Edit: let me explain better.
1) If $E^2$ is not empty, it means $\exists x$ such that $|f(x,y)-f(x,0)|=\epsilon, \forall y<\delta$. But if $f(x,\cdot)$ is continuous, this cannot happen.
2) This is more complicate than I thought. And David's hint is really important. Define $F_y(x)=f(x,y)-f(x,0)$. Then $A_y = F_y^{-1}(-\epsilon,\epsilon)$ is measurable. As David says, the definition of $E_{\epsilon,\delta}$ can be replaced by
$$ E_{\epsilon,\delta}=\{x:|F_y(x)|<\epsilon,\forall y<\delta, y\in\mathbb{Q}\} $$
Therefore, it follows that $E_{\epsilon,\delta}=\cap A_y$, where the intersection is over all $y<\delta,y\in\mathbb{Q}$. Since each $A_y$ is measurable and the intersection is countable, we obtain that $E_{\epsilon,\delta}$ is measurable.