∴C coordinate is (0,4),
Let y=0, then x=- 1,
∴B coordinate is (-1, 0),
∵ parabola y=ax2-2ax+c(a≠0) passes through points b and c,
∴0=a+2a+c4=c，
Solution: a =? 43c=4，
The analytical formula of parabola is y=-43x2+83x+4.
Let y=0, 0=-43x2+83x+4,
Solution: x=- 1 or 3,
∴A's coordinates are: (3,0);
(2) Let the analytical formula of straight line AC be y=kx+b,
∫ a (3,0), point c (0 0,4),
∴ 3k+b = 0b = 4, and k =? 43b=4，
The analytical formula of ∴ straight line AC is y =-43x+4.
The abscissa of point m is m, and point m is on AC.
The coordinate of point ∴M is (m, -43m+4),
∵ The abscissa of point P is m, and point P is on parabola y=-43x2+83x+4.
The coordinate of point p is (m, -43m2+83m+4).
∴pm=pe-me=(-43m2+83m+4)-(-43m+4)=-43m2+4m，
Namely pm =-43m2+4m (0 < m < 2);
(3) Under the condition of (2), connect PC, and there is such a point P in the parabolic part above CD, so that the triangle with P, C and F as vertices is similar to △AEM. The reason for this is the following：
AE=3-m，EM=-43m+4，CF=m，PF =-43m2+83m+4-4 =-43m2+83m。
If a triangle with vertices p, c and f is similar to △AEM, there are two situations:
① If △PFC∽△AEM, then PF: AE = FC: EM,
That is, (-43m2+83m): (3-m) = m: (-43m+4),
∵m≠0 and m≠3,
∴m=23 16.
② If △CFP∽△AEM, then CF: AE = PF: EM.
Namely m: (3-m) = (-43m2+83m): (-43m+4),
∵m≠0 and m≠3,
∴m= 1.
To sum up, there is such a point p that △PFC is similar to △AEM. At this time, the value of m is 23 16 or 1.