$解:(1)原式=\frac{2×\sqrt{3}}{3\sqrt{3}×\sqrt{3}}=\frac{2\sqrt{3}}{9}$
$(2)原式=\frac{2+\sqrt{3}}{(2-\sqrt{3})×(2+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{2}}{(\sqrt{3}-\sqrt{2})×(\sqrt{3}+\sqrt{2})}$
$=2+\sqrt{3}+\sqrt{3}+\sqrt{2}$
$=2+2\sqrt{3}+\sqrt{2}$
$(3)∵ \sqrt{2024}-\sqrt{2023}=\frac{1}{\sqrt{2024}+\sqrt{2023}}, \sqrt{2025}-\sqrt{2024}=\frac{1}{\sqrt{2025}+\sqrt{2024}},$
$而 \sqrt{2025}+ \sqrt{2024}>\sqrt{2024} +\sqrt{2023},$
$∴ \frac{1}{\sqrt{2025}+\sqrt{2024}}<\frac{1}{\sqrt{2024}+\sqrt{2023}} ,$
$即 \sqrt{2025}-\sqrt{2024}< \sqrt{2024}-\sqrt{2023}$