1.阅读下面的材料,并解答问题:$\frac{1}{{\sqrt{2}+1}}=\frac{{1×({\sqrt{2}-1})}}{{({\sqrt{2}+1})({\sqrt{2}-1})}}=\frac{{\sqrt{2}-1}}{{{{({\sqrt{2}})}^2}-{1^2}}}=\sqrt{2}$-1;$\frac{1}{{\sqrt{3}+\sqrt{2}}}=\frac{{1×({\sqrt{3}-\sqrt{2}})}}{{({\sqrt{3}+\sqrt{2}})({\sqrt{3}-\sqrt{2}})}}=\frac{{\sqrt{3}-\sqrt{2}}}{{{{({\sqrt{3}})}^2}-{{({\sqrt{2}})}^2}}}=\sqrt{3}-\sqrt{2}$;$\frac{1}{{\sqrt{4}+\sqrt{3}}}=\frac{{1×({\sqrt{4}-\sqrt{3}})}}{{({\sqrt{4}+\sqrt{3}})({\sqrt{4}-\sqrt{3}})}}=\frac{{\sqrt{4}-\sqrt{3}}}{{{{({\sqrt{4}})}^2}-{{({\sqrt{3}})}^2}}}=\sqrt{4}-\sqrt{3}$;…
(1)填空:$\frac{1}{{\sqrt{5}+\sqrt{4}}}$=$\sqrt{5}-2$,$\frac{1}{{\sqrt{2017}+\sqrt{2016}}}$=$\sqrt{2017}-12\sqrt{14}$;$\frac{1}{{\sqrt{n+1}+\sqrt{n}}}$=$\sqrt{n+1}-\sqrt{n}$(n为正整数);
(2)化简:$\frac{2}{{\sqrt{2}-1}}$.