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1. To maximize the number of elements in common between \(A\) and \(B\text{,}\) make \(A \subset B\text{.}\) This would give \(\card{A \cap B} = 10\text{.}\)

2. \(A\) and \(B\) might have no elements in common, giving \(\card{A\cap B} = 0\text{.}\)

3. \(15 \le \card{A \cup B} \le 25\text{.}\) In fact, when \(\card{A \cap B} = 0\) then \(\card{A \cup B} = 25\) and when \(\card{A \cap B} = 10\) then \(\card{A \cup B} = 15\text{.}\)