( 1)

Suppose that the side length of the cut square is xcm. According to the meaning of the question, we can get (40-2x) 2 = 484, and then we can get it.

② Assuming that the side length of the cut square is acm and the side area of the box is ycm^2, the functional relationship between Y and X is: y=4(40-2a)a, which can be obtained by using the maximum value of quadratic function;

(2) Assuming that the height of the cut rectangular frame is xcm and the surface area of the folded rectangular frame is 550cm^2, the equation can be obtained and solved.

Solution: Solution: (1)① Let the side length of the cut square be xcm. ..

Then (40-2x) 2 = 484,

I.e. 40-2x =+-22,

The solution is x 1=3 1 (irrelevant, omitted), x2=9,

The side length of the cut square is 9 cm.

② The lateral area has the maximum value.

Let the side length of the cut small square be acm and the side area of the box be ycm^2.

The functional relationship between y and a is: y=4(40-2a)a,

That is y =-8a 2+ 160a,

That is y =-8 (a- 10) 2+800,

When ∴a= 10 is the maximum, y = 800.

That is, when the side length of the cut square is 10cm, the maximum lateral area of the rectangular frame is 800cm^2. ..

(2) In the clipping diagram as shown in the figure, let the height of the clipped rectangular frame be tcm. ..

2(40-2t)(20t)+2x(20t)+2x(40-2t)= 550，

Solution: t 1=-35 (irrelevant, omitted), t2 = 15.

The height of the cut rectangular box is 15cm. ..

40-2× 15= 10 (cm),

20- 15=5 (cm),

At this time, the cuboid box is 10cm long and 5cm wide, 15cm. Tall.