1) $$7\sqrt{2}-3\sqrt{8}+4\sqrt{18}=7\sqrt{2}-3\sqrt{4\cdot 2}+4\sqrt{9\cdot 2}=7\sqrt{2}-3\cdot 2\sqrt{2}+4\cdot 3\sqrt{2}=7\sqrt{2}-6\sqrt{2}+12\sqrt{2}=(7-6+12)\sqrt{2}=13\sqrt{2}$$
2) $$(\sqrt{90}-\sqrt{40})\sqrt{10}=(\sqrt{9\cdot 10}-\sqrt{4\cdot 10})\sqrt{10}=(3\sqrt{10}-2\sqrt{10})\sqrt{10}=\sqrt{10}\cdot \sqrt{10}=10$$
3) $$(3\sqrt{5}-2)^2=(3\sqrt{5})^2-2\cdot 3\sqrt{5}\cdot 2 + 2^2=9\cdot 5-12\sqrt{5}+4=45+4-12\sqrt{5}=49-12\sqrt{5}$$
4) $$(2\sqrt{3}+3\sqrt{5})(2\sqrt{3}-3\sqrt{5})=(2\sqrt{3})^2-(3\sqrt{5})^2=4\cdot 3 - 9\cdot 5=12-45=-33$$
Ответ: 1) $$13\sqrt{2}$$; 2) 10; 3) $$49-12\sqrt{5}$$; 4) -33