2.【问题分析】
在解决问题“已知$a=\frac{1}{\sqrt{2}-1}$,求$3a^2-6a-1$的值”时,小李是这样分析与解答的:
$\because a=\frac{1}{\sqrt{2}-1}=$$=\sqrt{2}+1$,
$\therefore a-1=\sqrt{2}$,
$\therefore (a-1)^2=2$,即$a^2-2a+1=2$,
$\therefore a^2-2a=1$,
$\therefore 3a^2-6a=3$,
$\therefore 3a^2-6a-1=2$.
【学以致用】
请你根据小李的分析过程,解决下列问题:
(1)化简:$\frac{2}{3-\sqrt{7}}$.
(2)若$a=\frac{1}{3+2\sqrt{2}}$,求$3a^2-18a+1$的值.
【拓展延伸】
(3)计算:$(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\dots+\frac{1}{\sqrt{2025}+\sqrt{2026}})×(\sqrt{2026}+1)$.

2.解:(1)$\dfrac{2}{3-\sqrt{7}}=\dfrac{2(3+\sqrt{7})}{(3-\sqrt{7})(3+\sqrt{7})}=\dfrac{2(3+\sqrt{7})}{9-7}=\dfrac{2(3+\sqrt{7})}{2}=3+\sqrt{7}$.
(2)$a=\dfrac{1}{3+2\sqrt{2}}=\dfrac{3-2\sqrt{2}}{(3+2\sqrt{2})(3-2\sqrt{2})}=\dfrac{3-2\sqrt{2}}{9-8}=3-2\sqrt{2}$, $\therefore a-3=-2\sqrt{2}$,
$\therefore (a - 3)^2=(-2\sqrt{2})^2=8$,即$a^2-6a+9=8$,
$\therefore a^2-6a=-1$, $\therefore 3a^2-18a=-3$, $\therefore 3a^2-18a+1=-3+1=-2$.
(3)原式=$(-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-\dots-\sqrt{2025} + \sqrt{2026}) × (\sqrt{2026} + 1) = (\sqrt{2026}-1)×(\sqrt{2026}+1)=2026-1=2025$.