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Ответ:   4 .

[tex]\displaystyle \frac{4}{\sqrt4+\sqrt8}+\frac{4}{\sqrt8+\sqrt{12}}+\frac{4}{\sqrt{12}+\sqrt{16}}+...+\frac{4}{\sqrt{32}+\sqrt{36}}=[/tex]  

Домножим дроби на выражения, сопряжённые знаменателям, и вынесем 4 за скобку . Потом воспользуемся формулой разности квадратов:   [tex]\bf (a+b)(a-b)=a^2-b^2[/tex]  .

[tex]\displaystyle =4\cdot \Big(\frac{\sqrt4-\sqrt8}{(\sqrt4+\sqrt8)(\sqrt4-\sqrt8)}+\frac{\sqrt8-\sqrt{12}}{(\sqrt8+\sqrt{12})(\sqrt8-\sqrt{12})}+\\\\\\+\frac{\sqrt{12}-\sqrt{16}}{(\sqrt{12}+\sqrt{16})(\sqrt{12}-\sqrt{16})}+...+\frac{\sqrt{32}-\sqrt{36}}{(\sqrt{32}+\sqrt{36})({\sqrt{32}-\sqrt{36})}}\Big)=[/tex]  

[tex]\displaystyle =4\cdot \Big(\dfrac{\sqrt4-\sqrt8}{4-8}+\dfrac{\sqrt8-\sqrt{12}}{8-12}+\frac{\sqrt{12}-\sqrt{16}}{12-16}+...+\frac{\sqrt{32}-\sqrt{36}}{32-36}}\Big)=\\\\\\=4\cdot \Big(\dfrac{\sqrt4-\sqrt8}{-4}+\dfrac{\sqrt8-\sqrt{12}}{-4}+\frac{\sqrt{12}-\sqrt{16}}{-4}+...+\frac{\sqrt{32}-\sqrt{36}}{-4}}\Big)=[/tex]

[tex]=-\Big(\sqrt4\underbrace{-\sqrt8+\sqrt8}_{0}\underbrace{-\sqrt{12}+\sqrt{12}}_{0}-\sqrt{16}\ +\ ...+\sqrt{28}\underbrace{-\sqrt{32}+\sqrt{32}}_{0}-\sqrt{36}\Big)=\\\\\\=-\sqrt4+\sqrt{36}=6-2=\bf 4[/tex]