а) $$2\sqrt{2}-\sqrt{50}-3\sqrt{98}=2\sqrt{2}-\sqrt{25\cdot2}-3\sqrt{49\cdot2}=2\sqrt{2}-5\sqrt{2}-3\cdot7\sqrt{2}=2\sqrt{2}-5\sqrt{2}-21\sqrt{2}=-24\sqrt{2}$$
б) $$(\sqrt{3}+\sqrt{2})^2=(\sqrt{3})^2+2\cdot\sqrt{3}\cdot\sqrt{2}+(\sqrt{2})^2=3+2\sqrt{6}+2=5+2\sqrt{6}$$
Ответ: а) $$-24\sqrt{2}$$; б) $$5+2\sqrt{6}$$