V father = 400 ÷ 4 =100m/min;

V female = 200 ÷ 4 = 50m/min;

(2) The daughter of the image will meet her father three times.

Let the analytical formula of DG be y 1=k 1x+b 1, and the analytical formula of EF be y2=k2x+b2. From the image, we can get

200 = 5k 1+b 10 = 9k 1+b 1，200=6k2+b20=8k2+b2，

Solution: k 1 =? 50b 1=450，k2=？ 100b2=800，

∴y 1=-50x+450,y2=- 100x+800，

When y 1=y2, -50x+450=- 100x+800,

x=7，

∴y= 100

On the daughter's way back and forth, the last time the father and daughter met was at a distance of 100 meters.