$解:(2)②原式=(1-\frac{1}{2})×(1+\frac{1}{2})×(1-\frac{1}{3})×(1+\frac{1}{3})×...×(1-\frac{1}{2024})×(1+\frac{1}{2024})$
$=\frac{1}{2}×\frac{3}{2}×\frac{2}{3}×\frac{4}{3}×...×\frac{2023}{2024}×\frac{2025}{2024}$
$=\frac{1}{2}×\frac{2025}{2024}$
$=\frac{2025}{4048}$