【答案解析】(1) Let n,n + 2, and n + 4 be the consecutive even integers. Using the Pythagorean theorem, we have n2 + (n + 1)2 = (n + 4)2. Because this is a quadratic equation that may have two solutions, we need to investigate further to determine whether there is a unique hypotenuse length.n2 + (n + 2)2=(n+4)2= (n + 4)2n2 + n2 + 4n + 4=n2+8n+16n2 - 4n - 12 = 0(n-6)(n + 2) = 0Therefore, n = 6 or n = -2. Since n = -2 corresponds to side lengths of-2,0, and 2, we discard n = -2. Therefore n = 6, the hypotenuse has length n + 4 = 10; SUFFICIENT.(2) Let the side lengths be a, b, and a + 4. Using the Pythagorean theorem, we have a2 + b2 = (a + 4)2. Expanding and solving for b in terms of a will facilitate our search for multiple hypotenuse length possibilities.a2 + b2 = (a + 4)2a2+ b2 = a2 + 8a+16b2= 8a+16b =

When a = 1, we obtain side lengths 1 and

, and hypotenuse length 5. When a = 2, we obtainside lengths 2 and
