$a)$  

$\frac{1}{3}\sqrt{6}+1\frac{2}{3}\sqrt{6}=\frac{\sqrt{6}}{3}+\frac{5}{3}\sqrt{6}=\frac{\sqrt{6}}{3}+\frac{5\sqrt{6}}{3}=\frac{6\sqrt{6}}{3}=2\sqrt{6}$  

 

$b)$  

$\sqrt{7}-3\frac{2}{3}\sqrt{7}=\sqrt{7}-\frac{11}{3}\sqrt{7}=\left(1-\frac{11}{3}\right)\cdot \sqrt{7}=\left(\frac{3}{3}-\frac{11}{3}\right)\cdot \sqrt{7}=-\frac{8}{3}\sqrt{7}=-2\frac{2}{3}\sqrt{7}$  

 

$c)$  

$3\sqrt{11}-\frac{\sqrt{11}}{3}=\frac{9\sqrt{11}}{3}-\frac{\sqrt{11}}{3}=\frac{8\sqrt{11}}{3}=\frac{8}{3}\sqrt{11}=2\frac{2}{3}\sqrt{11}$ $3\sqrt{11}-\frac{\sqrt{11}}{3}=\frac{9\sqrt{11}}{3}-\frac{\sqrt{11}}{3}=\frac{8\sqrt{11}}{3}=\frac{8}{3}\sqrt{11}=2\frac{2}{3}\sqrt{11}$  

 

$d)$  

$\frac{\sqrt{2}}{5}+\frac{\sqrt{2}}{10}=\frac{2\sqrt{2}}{10}+\frac{\sqrt{2}}{10}=\frac{3\sqrt{2}}{10}=\frac{3}{10}\sqrt{2}=0,3\sqrt{2}$  

 

$e)$  

$-6\sqrt{3}\cdot \frac{1}{2}=-\frac{6\sqrt{3}}{2}=-3\sqrt{3}$  

 

$f)$  

$5\sqrt{13}\cdot \left(-0,4\right)=5\sqrt{13}\cdot \left(-\frac{4}{10}\right)=-\frac{20\sqrt{13}}{10}=-2\sqrt{13}$  

 

$g)$  

$0,1\sqrt{2}\cdot \sqrt{2}=0,1\cdot {\left(\sqrt{2}\right)}^2=0,1\cdot 2=0,2$  

 

$h)$  

$1,2\sqrt{10}\cdot 2\sqrt{10}=1,2\cdot 2\cdot {\left(\sqrt{10}\right)}^2=1,2\cdot 2\cdot 10=24$    

 

$i)$  

$\frac{5}{6}\sqrt{3}\cdot \frac{\sqrt{3}}{2}=\frac{5\sqrt{3}}{6}\cdot \frac{\sqrt{3}}{2}=\frac{5\cdot {\left(\sqrt{3}\right)}^2}{12}=\frac{5\cdot 3}{12}=\frac{15}{12}=\frac{5}{4}=1\frac{1}{4}$   

 

$j)$  

$\frac{5\sqrt{7}-10\sqrt{3}}{5}=\frac{5\cdot \left(\sqrt{7}-2\sqrt{3}\right)}{5}=\sqrt{7}-2\sqrt{3}$   

 

$k)$  

${\left(4\sqrt{3}\right)}^2+{\left(\sqrt{2}\right)}^2=4^2\cdot {\left(\sqrt{3}\right)}^2+{\left(\sqrt{2}\right)}^2=16\cdot 3+2=48+2=50$    

 

$l)$  

$\frac{{\left(3\sqrt{5}\right)}^2}{{\left(5\sqrt{3}\right)}^2}=\frac{3^2\cdot {\left(\sqrt{5}\right)}^2}{5^2\cdot {\left(\sqrt{3}\right)}^2}=\frac{3^2\cdot 5}{5^2\cdot 3}=\frac{3^{\sout{2}}\cdot \sout{5}}{5^{\sout{2}}\cdot \sout{3}}=\frac{3}{5}=0,6$