I would like to express those text ( Original text is in french ) into mathematical form
$1$. The map $$Z\longmapsto \bar{A} \cup Z $$ is a bijection of the set of All the parts $Z$ of $A$ on the set of the parts $X$ of $E$ Such that $X \cup A = E$.
In mathematical form : \begin{align*} 1: \{ \mathcal{P}\left(A\right)\} &\rightarrow \{\mathcal{P}\left(X \right)\mid X\subset E: X\cup A=E\}\\ Z&\mapsto \overline{A}\cup Z \end{align*}
$2$.The map $$Z\longmapsto (A\cap B ) \cup Z $$ is a bijection of the set of parts of $(A\cup B )\setminus (A\cap B)$ onto the set of parts X of E such that $A\cap B \subset X \subset A\cup B$.
In mathematical form : \begin{align*} 2: \{ \mathcal{P}\left((A\cup B )\setminus (A\cap B)\right)\} &\rightarrow \{\mathcal{P}\left(X \right)\mid X\subset E: A\cap B \subset X \subset A\cup B\}\\ Z&\mapsto (A\cap B ) \cup Z \end{align*}
Original text in french :

It appears that the sets $A$, $B$, and $E$ are given. So, I think it is better to define two new sets : $$\mathcal{C}=\{ X\in \mathcal{P}(E)| X\cup A=E \}$$ and
$$\mathcal{D}=\{ X\in \mathcal{P}(E)| A\cap B \subset X \subset A\cup B \}$$
Then the two maps :
\begin{align*} \mathcal{P}\left(A\right) \rightarrow \mathcal{C} \\ Z \mapsto \overline{A}\cup Z \end{align*} and
\begin{align*} \mathcal{P}\left((A\cup B )\setminus (A\cap B)\right) &\rightarrow \mathcal{D}\\ Z&\mapsto (A\cap B ) \cup Z \end{align*} are bijective.