( 1)4×2×2= 16 cm2

6× 2 =12cm

(3) (20-12) ÷ 2+6 = 4+6 =10 second

? 12 ÷ 2+10 = 6+10 =16 seconds.

Answer: (1) After moving for 4 seconds, the area of the overlapping part is 16 cm2; (2) The side length of a square is12 cm; (3) Fill in10s in the first bracket and16s in the second bracket.

Interpretation of the meaning of the question:

(1) The speed of the paper strip is 2cm/s, and when it runs for 4 seconds, the distance traveled by the paper strip is 4× 2 = 8cm, so the graphic length of the overlapping part is 8cm, and the known width is 2cm, so the area is 8× 2 =16 cm2. (as shown in the figure below)

(2) As can be seen from the figure below, it takes 6 seconds for a * * * to travel from the time when the note enters the square to the time when the overlapping area between the note and the square reaches the maximum (as shown in Figure 2).

So the side length of a square is 6×2= 12 cm.

(3) When the overlapping area between the paper strip and the square is the largest, as shown in the following figure, the length of the left side of the paper strip that does not enter the square is 20- 12 = 8cm, and after the paper strip walks through this 8cm, the overlapping area begins to decrease, so the time for keeping the overlapping area unchanged is 8÷2=4 seconds, and the first time is 6+4 = 60 seconds.

When the overlapping area of the note begins to decrease, the time is 10 second, and the distance from the overlapping area to the square is the side length of the square, so the time of walking will be 12÷2=6 seconds, and the second time will be 10+6= 16 seconds.