\((\subseteq)\text{:}\) Let \(x\in A\cup(B\cap C)\text{.}\) Then \(x\in A\) or \(x\in B\cap C\text{.}\)
Case 1: \(x\in A\text{.}\) Then \(x\in A\cup B\) by the second subset relation. Similarly, \(x\in A\cup C\text{.}\) Thus, \(x\in A\cup B\) and \(x\in A\cup C\text{.}\)
Therefore, \(x\in (A\cup B) \cap (A\cup C)\text{.}\)
Case 2: \(x\in B\cap C\text{.}\) Then \(x\in B\) and \(x\in C\text{.}\) Then \(x\in A\cup B\) and \(x\in A\cup C\) by the second subset relation. Thus, \(x\in A\cup B\) and \(x\in A\cup C\text{.}\)
Therefore, \(x\in (A\cup B) \cap (A\cup C)\text{.}\)
\((\supseteq)\text{:}\) Let \(x\in (A\cup B)\cap (A\cup C)\text{.}\) Then \(x\in A\cup B\) and \(x\in A\cup C\text{.}\)
Case 1: \(x\in A\text{.}\) Then \(x\in A\cup (B\cap C)\) by the second subset relation (since we can do the union with any set).
Case 2: \(x\notin A\text{.}\) Since \(x\in A\cup B\) by assumption, \(x\in A\) or \(x\in B\text{.}\) Since \(x\notin A\text{,}\) \(x\in B\text{.}\)
Similarly, since \(x\in A\cup C\text{,}\) and \(x\notin A\text{,}\) \(x\in C\text{.}\) Thus, \(x\in B\cap C\text{.}\)
Therefore, \(x\in A\cup (B\cap C)\) (again by the second subset relation).