I SPOSÓB

$a)3\sqrt{2}=\sqrt{3^2}\cdot \sqrt{2}=\sqrt{9}\cdot \sqrt{2}=\sqrt{9\cdot 2}=\sqrt{18}$

$2\sqrt{3}=\sqrt{2^2}\cdot \sqrt{3}=\sqrt{4}\cdot \sqrt{3}=\sqrt{4\cdot 3}=\sqrt{12}$

$3\sqrt{2}>2\sqrt{3}$

$b)\frac{\sqrt{2}}{2}=\frac{1}{2}\cdot \sqrt{2}=\sqrt{{\left(\frac{1}{2}\right)}^2}\cdot \sqrt{2}=\sqrt{\frac{1}{4}}\cdot \sqrt{2}=\sqrt{\frac{1}{4}\cdot 2}=\sqrt{\frac{2}{4}}=\sqrt{\frac{1}{2}}$

$\frac{\sqrt{3}}{3}=\frac{1}{3}\cdot \sqrt{3}=\sqrt{{\left(\frac{1}{3}\right)}^2}\cdot \sqrt{3}=\sqrt{\frac{1}{9}}\cdot \sqrt{3}=\sqrt{\frac{1}{9}\cdot 3}=\sqrt{\frac{3}{9}}=\sqrt{\frac{1}{3}}$