(1)证明:设A(x1,y1),B(x2,y2),A、B的中点为P(a,b),由已知得y1^2-y2^2=2px1-2px2,
所以(y1+y2)(y1-y2)=2p(x1-x2),直线AB的斜率为(y1-y2)/(x1-x2)=2p/(y1+y2)=p/b,直线AB的
垂直平分线的斜率为-b/p,又因为|AF|+||BF|=2|MF|,所以(x1+p/2)+(x2+p/2)=2(x0+p/2),即
x1+x2=2x0,P(x0,b).直线AB的垂直平分线方程为y-b=-(b/p)(x-x0),即b(x-x0-p)+py=0,所以线段AB的垂直平分线过定点Q(x0+p,0);
(2)由已知得x0+p/2=4,x0+p=6,解得p=4,所以抛物线方程为y^2=8x.