In the book of Mathematical Analysis by Zorich, at page 123, it is asked that
For question 1(b), I can give the set $$A = \{x \in \mathbb{Q}| 0\leq x \leq 1\}.$$
and for question 1(a), I have argued that if $I = \{I_i\}$ is an at most countable open cover for $E$, than $\bar E \subset \bar I$, and $\mu(\bar I_i) \leq \mu(I_i) + \mu(\partial I_i)$ and for any interval $I_i \in \mathbb{R}^n$, $\partial I_i$ is has measure zero, hence $\bar E$ has to have measure zero.
However, how does these two results do not conflict with each other ?
Edit:
As it is pointed out in the comments by @Dap, $\mu$ represents the Jordan measure, so the problem has been solved thanks to the comments.
"Mathematical Analysis II" uses $\mu$ for Jordan measure, which is not defined for your set $A.$ And while the book defines Lebesgue measure zero sets, it never defines or uses Lebesgue measure.
To answer 1(a), use a finite cover $C_1,\dots,C_k$ of $E$ by $n$-dimensional closed intervals of total measure less than $\epsilon,$ and note that the closure of $E$ is also covered by the same set. (If Jordan measure is defined in terms of half-open intervals or whatever, you might need to enlarge them slightly while keeping the total measure less than $2\epsilon$ say.)
Your answer for 1(b) is correct, as long as you know why that set is a Lebesgue measure zero set.