\(1,2\sqrt{3}+\sqrt{27}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}=\sqrt{75}>\sqrt{74}\\ 2,\left(3+\sqrt{5}\right)^2=14+6\sqrt{5}\\ \left(2\sqrt{2}+\sqrt{6}\right)^2=14+4\sqrt{3}\\ 6\sqrt{5}=\sqrt{180}>\sqrt{48}=4\sqrt{3}\\ \Rightarrow3+\sqrt{5}>2\sqrt{2}+\sqrt{6}\\ 3,3\sqrt{3}+\sqrt{26}+1>3\sqrt{3}+\sqrt{3}\left(\sqrt{26}>\sqrt{3}\right)\\ \Rightarrow\sqrt{27}+\sqrt{26}+1>4\sqrt{3}=\sqrt{48}\\ 4,\dfrac{1}{\sqrt{15}+\sqrt{14}}< \dfrac{1}{\sqrt{14}+\sqrt{13}}\left(\sqrt{15}+\sqrt{14}>\sqrt{14}+\sqrt{13}\right)\\ \Rightarrow\sqrt{15}-\sqrt{14}< \sqrt{14}-\sqrt{13}\)