\(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}-\dfrac{3-\sqrt{2}}{3+\sqrt{2}}\\ =\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}-\dfrac{\left(3-\sqrt{2}\right)^2}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}\\ =\dfrac{3-2\sqrt{2}}{2-1}-\dfrac{11-6\sqrt{2}}{9-2}\\ =3-2\sqrt{2}-\dfrac{11-6\sqrt{2}}{7}\\ =\dfrac{7\left(3-2\sqrt{2}\right)-\left(11-6\sqrt{2}\right)}{7}\\ =\dfrac{21-14\sqrt{2}-11+6\sqrt{2}}{7}\\ =\dfrac{10-8\sqrt{2}}{7}\)
\(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\\ =\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{3\sqrt{2}+2\sqrt{3}}\\ =\dfrac{3\sqrt{2}+2\sqrt{3}-3\sqrt{2}+3\sqrt{3}}{\left(\sqrt{2}-\sqrt{3}\right)\left[\sqrt{6}\left(\sqrt{2}+\sqrt{3}\right)\right]}\\ =\dfrac{5\sqrt{3}}{\sqrt{6}\left(3-2\right)}\\ =\dfrac{5\sqrt{3}}{\sqrt{2}.\sqrt{3}}\\ =\dfrac{5}{\sqrt{2}}\)
