If $\mu(E)=0$, show that $\mu(E\cup A)=\mu(A\setminus E)=\mu(A)$. 
If $\mu(E)=0$, show that $\mu(E\cup A)=\mu(A\setminus E)=\mu(A)$.

I just started learning about measure this week, so I don't know any theory about measure except the definition of outer measure and that a countable set has measure zero.
Remark: $\mu(\cdot)$ denoted the outer measure
Attempt.
Suppose $\mu(E)=0$. Then $\mu(A\cup E)\leq\mu(A)+\mu(E)=\mu(A)$.
If I can also show that $\mu(A \cup E)\geq \mu (A)$, then equality would follow.
I know that if $A$ and $E$ were disjoint compact sets, then $\mu(A)+\mu(E)\leq \mu(A \cup E)$, so maybe I can say something like a compact set in $\mathbb{R}$ is closed and bounded by Heine-Borel, and somehow use the fact that $\mu([a,b])=\mu((a,b))$ to help.
For $\mu(A\setminus E)=\mu(A)-\mu(E)=\mu(A)$
A hint in the right direction would be appreciated. Thanks.

Note that if $S \subset T$, then $\mu(S) \leq \mu(T)$ provided both $S$ and $T$ are measurable. So, $\mu(A) \leq \mu(A \cup E)$, and you already have the other inequality.
Similar idea will work for the next equality as well.
