For the first claim, we have
\begin{equation}
\begin{split}
E \cap (F \setminus E)
& = \{ x | x \in E \text{ and } x \in F \setminus E \} \\
& = \{ x | x \in E \text{ and } x \in F \text{ and } x \not\in E \}
= \{ x \in F | x \in E \text{ and } x \not\in E \}
= \emptyset
\end{split}
\end{equation}
For the second claim, if $x \in E \cup F$, then $x \in E$, or $x \in F$, or both. If $x \in E$, then $x \in E \cup (F \setminus E)$ irrespective of whether $x \in F$ or $x \not\in F$ since $E$ is part of the union. If $x \in F$, but $x \not\in E$ then $x \in F \setminus E$ and therefore $x \in E \cup (F \setminus E)$ since $F \setminus E$ is part of the union. Since $x \in E \cup F$ was arbitrary, we have the inclusion $E \cup F \subseteq E \cup (F \setminus E)$.
In the other direction, if $x \in E \cup (F \setminus E)$, then $x \in E$ or $x \in F \setminus E$. The first claim showed that $E$ and $F \setminus E$ are disjoint, so we can't have the situation where $x$ belongs to both sets. If $x \in E$, then $x \in E \cup F$ since $E$ is part of the union. If $x \in F \setminus E$, then $x \in F$ and therefore $x \in E \cup F$ since $F$ is part of the union. Since $x \in E \cup (F \setminus E)$ was arbitrary, we have the inclusion $E \cup (F \setminus E) \subseteq E \cup F$.
Since we have inclusion in both directions, it follows that $E \cup F = E \cup (F \setminus E)$. $\square$
