2．
在进行二次根式化简时，我们有时会碰上如\(\dfrac{5}{\sqrt{3}}\)，\(\sqrt{\dfrac{2}{3}}\)，\(\dfrac{2}{\sqrt{3}+1}\)一样的式子\(.\)其实我们还可以将其进一步化简：\(\dfrac{5}{\sqrt{3}}=\dfrac{5\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}=\dfrac{5}{3}\sqrt{3}(\)一\()\)，\(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{2\times 3}{3\times 3}}=\dfrac{\sqrt{6}}{3}(\)二\()\)，\(\dfrac{2}{\sqrt{3}+1}=\dfrac{2\times (\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\dfrac{2(\sqrt{3}-1)}{{{(\sqrt{3})}^{2}}-{{1}^{2}}}=\sqrt{3}-1(\)三\()\)，\(\dfrac{2}{\sqrt{3}+1}\)还可以用下面方法化简：\(\dfrac{2}{\sqrt{3}+1}=\dfrac{3-1}{\sqrt{3}+1}=\dfrac{{{(\sqrt{3})}^{2}}-{{1}^{2}}}{\sqrt{3}+1}=\dfrac{(\sqrt{3}+1)(\sqrt{3}-1)}{\sqrt{3}+1}=\sqrt{3}-1(\)四\()\)以上这种化简的方法叫做分母有理化．
\((1)\)请化简\(\dfrac{2}{\sqrt{5}+\sqrt{3}}=\_\_\_\_\_\_\_\_\)；
\((2)\)若\(a\)是\(\sqrt{2}\)的小数部分，则\(\dfrac{3}{a}=\_\_\_\_\_\_\_\_\)；
\((3)\)矩形的面积为\(3\sqrt{5}+1\)，一边长为\(\sqrt{5}-2\)，则它的周长为________；
\((4)\)化简\(\dfrac{2}{1+\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{9}}+\dfrac{2}{\sqrt{9}+\sqrt{13}}+\cdots +\dfrac{2}{\sqrt{4n-3}+\sqrt{4n+1}}\)．