1. solution: let x be the distance and y be the fare, then Y A = {Y ={y=6, and x ≤ 3 km; Y = 6+ 1.5x，x > 3km}，y ={y= 10，xx≤3km； Y = 10+ 1.2x, x > 3km}, when x > 3km, Y A -y B =-4+0.3x, where when x=40/3, the price difference between Party A and Party B is 0, and when x is between 3km and 40/3km, the price difference is.

2. solution: when the distance is 10km, x= 10-3=7km, and it is substituted into Y A -y B =-4+0.3x =- 1.9 < 0, which means that Y A < Y B is1.