(x+2)？ ≥0; (y-3)？ ≥0; For the equation to be established, we only need to solve x+2=0 and y-3=0, and we can get x=-2 and y=3.

xy=-6

2，(3a？ b？ c^4)？ □ 1/3a？ b^4)=9a^4b^4c^8/(- 1/3a？ b^4)=-27a？ c^8

3,(2+ 1)(2? + 1)(2^4+ 1)(2^8+ 1)(2^ 16+ 1)/(2^32- 1)

=(2- 1)(2+ 1)(2? + 1)(2^4+ 1)(2^8+ 1)(2^ 16+ 1)/(2^32- 1)

=(2? - 1)(2? + 1)(2^4+ 1)(2^8+ 1)(2^ 16+ 1)/(2^32- 1)

=(2^4- 1)(2^4+ 1)(2^8+ 1)(2^ 16+ 1)/(2^32- 1)

=(2^8- 1)(2^8+ 1)(2^ 16+ 1)/(2^32- 1)

=(2^ 16- 1)(2^ 16+ 1)/(2^32- 1)

=(2^32- 1)/(2^32- 1)

= 1