(i) $A \cup (A \cap B) = A$

To prove $A \cup (A \cap B) = A$, we show $A \subseteq A \cup (A \cap B)$ and $A \cup (A \cap B) \subseteq A$.

Let $x \in A$. By definition of union, $x \in A \cup (A \cap B)$. Thus $A \subseteq A \cup (A \cap B)$.

Let $x \in A \cup (A \cap B)$. Then $x \in A$ or $x \in (A \cap B)$. If $x \in A$, then $x \in A$. If $x \in (A \cap B)$, then $x \in A$ and $x \in B$, so $x \in A$. Thus $A \cup (A \cap B) \subseteq A$.

Since both conditions hold, $\therefore \; \mathbf{A \cup (A \cap B) = A}$.

(ii) $A \cap (A \cup B) = A$

To prove $A \cap (A \cup B) = A$, we show $A \subseteq A \cap (A \cup B)$ and $A \cap (A \cup B) \subseteq A$.

Let $x \in A$. Since $A \subseteq A \cup B$, we have $x \in A \cup B$. Thus $x \in A$ and $x \in A \cup B$, so $x \in A \cap (A \cup B)$. Therefore $A \subseteq A \cap (A \cup B)$.

Let $x \in A \cap (A \cup B)$. By definition of intersection, $x \in A$ and $x \in A \cup B$. Thus $x \in A$. Therefore $A \cap (A \cup B) \subseteq A$.

Since both conditions hold, $\therefore \; \mathbf{A \cap (A \cup B) = A}$.