Solution: 1 ① See the figure below. The parameter equation of circle O is: x=3cost, y = 3sint.

② Mx=(4+3cost)/2=2+(3/2).....( 1)，My=5+3sint/2=(5/2)+(3/2)sint....(2)

The trajectory equation of point M is: (x-2) 2+(y-5/2) 2 = (3/2) 2;

2.(x- 1)^2+(y- 1)^2=x^2+y^2-2x-2y+2=4； 2(x+y)=x^2+y^2-2；

(x+y)=( 1/2)(x^2+y^2)- 1<； = xy- 1； ? Because (x- 1) 2+(y- 1) 2 = 4, when (x- 1) 2 = (y- 1) 2 = 2,

X- 1=y- 1=√2，x=y= 1+√2，

(x+y)max=( 1+√2)^2-2=3+2√2- 1= 1+2√2。