（1）由已知图象得：甲的速度为100km/h,乙的速度为50km/h,
答：甲、乙两车的速度分别为100km/h,50km/h．
（2）设乙车从B地返回到C地的函数解析式是y=kx+b
∵图象经过（5,0）,（9,200）两点）．
∴5k+b=09k+b=200
解得：$left{begin{array}{l}{k=50}\{b=-250}end{array}right.$,
∴y=50x-250,
答：乙车从B地返回到C地的过程中,y与x之间的函数关系式为y=50x-250．
（3）设甲车从A地到B地的函数解析式是y1=k1x+b1,
∵图象经过（0,600）,（6,0）两点,
∴$left{begin{array}{l}{600={b}_{1}}\{0=6{k}_{1}+{b}_{1}}end{array}right.$,解得：$left{begin{array}{l}{{k}_{1}=-100}\{{b}_{1}=;600}end{array}right.$,∴y1=-100x+600,
设甲车从B地到C地的函数解析式是y2=k2x+b2,
∵图象经过（8,200）,（6,0）两点,
∴$left{begin{array}{l}{0=6{k}_{2}+{b}_{2}}\{200=8{k}_{2}+{b}_{2}}end{array}right.$,解得：$left{begin{array}{l}{{k}_{2}=100}\{{b}_{2}=-;600}end{array}right.$,∴y2=100x-600,
由$left{begin{array}{l}{y=50x-250}\{{y}_{1}=-100x+600}end{array}right.$和$left{begin{array}{l}{y=50x-250}\{{y}_{2}=100x-600}end{array}right.$,
解得：y=$frac{100}{3}$（千米）和y=100（千米）．