(1) The straight line passes through the center m (4,2). As long as the central coordinate is substituted into the linear equation, the intercept b on the Y axis can be obtained, and b =10 can be obtained from 2=-2*4+b;

When the straight line is tangent to the circle, there is only one intersection point between them, so the quadratic equation obtained by substituting the straight line equation into the circle equation has only one root;

Circle: (x-4) 2+(y-2) 2 = 2 2, straight line: y=-2x+b, and the latter is substituted for the former: (x-4) 2+(-2x+b-2) 2 = 4;

Finishing: 5x2-4bx+(b-2) 2+12 = 0;

If this equation has only one real root, the discriminant of the root must be equal to 0: (4b) 2-4 * 5 * [(b-2) 2+12] = 0, and the solution is b = 5 √ 5;

When b=5+√5 (tangent to the right side) and b=5-√5 (tangent to the left side), the straight line y=-2x+b is tangent to the circle;

(2) When the straight line is on the left side of corner A, s = 0；; Substituting A (2,0) into the linear equation can get b=4, that is, when b≤4, S = 0；;

When a straight line sweeps across the rectangle from A to right, the area S linearly increases from 0 to 2*2/2=2 before crossing point D. Substituting the coordinates of d (2,2) into a straight line, we can get b, 2=-2*2+b, and b=6, that is, when 4.

When the straight line sweeps across the rectangle from D to right, the area S linearly increases from 2 to 4*2-2=6 before passing through B. Substituting the coordinates of B (6,0) into the linear equation, we can get B: 0 =-2 * 6+B, and B = 12. That is, when 6

When the straight line continues to sweep the rectangle from B to the right, the area S linearly increases from 6 to the maximum value of 4*2=8. Substituting the c (6,2) coordinate into the linear equation, we can get B: 2 =-2 * 6+B, and B = 14. That is when 12