Then as follows

There seems to be little information about this problem. I don't know the speed at which Xiao Lin walks, the speed at which his father drives, and the departure time every day. But after clever calculation, it is not difficult to get the answer.

Let's imagine that if my father meets Xiaolin halfway, continues to drive to the school gate, then goes back to the place where they met, and then takes Xiaolin home, then the time to get home will be the same as usual. So 10 minutes in advance is exactly equal to the time for my father to go back and forth from the place where they met to the school gate 1 time, that is, it takes five minutes for my father to walk Kobayashi by motorcycle (from the place where they met at the school gate). As my father still intends to arrive at the school gate at 6 o'clock sharp as usual, my father and son met at 5: 55 on the way. Because Kobayashi finishes school at 5: 30 every day, I walked for 25 minutes when I met him.

The question about post- 1 can be considered as follows: According to the above analysis, the time for father and son to meet is 5: 55. At this time, Kobayashi had left for 15 minutes, so Kobayashi left the school gate at 5: 40.

Hope to adopt!