Solution: (1) Because the image passes through point C, it can be obtained by bringing the coordinates of point C into a quadratic function: c= 1.

(2) Because the quadratic function has two different intersections with the X axis, the quadratic equation AX 2+BX+1= 0 corresponding to the quadratic function has two different roots. Therefore, according to Vieta's theorem, we can get: b 2-4ac = b 2-4a > 0-1

Since the coordinate of point A is known as (1, 0), a+b+c=a+b+ 1=0.

So b =-(a+1); Bring in the above formula 1:

(a+ 1)^2-4a>； 0

a^2+2a+ 1-4a=a^2-2a+ 1=(a- 1)^2>； 0 so when a is not equal to 1. Because a > has been marked in the stem; 0

So the range of a is: (0, 1) or (1, +∞).

(3) We should draw a sketch first and sketch out the basic details. The quadratic function passes through A( 1, 0), C(0, 1) and A >；; 0 The positive semi-axis symmetrical in the X direction can be drawn with the opening upward;

According to the symmetry of quadratic function, it is known that the quadrilateral ABCD formed by the intersection points A, B, C and D of X axis and y= 1 is an isosceles trapezoid.

Let the two roots of quadratic equation AX 2+BX+1= 0 be x 1, x2; Then x 1+x2=-b/a, x1x2 = c/a;

According to the square of | ab | = x1-x2 | = √ (x1-x2) 2, the root number value remains unchanged.

= √ (x 1+x2) 2-4x 1x2 To convert the square of the difference into the square of the sum, you need to subtract the product of 4 times.

= √ (b 2-4a)/a 2 brings in b=-(a+ 1) again.

=√(a- 1)^2/a^2

In this case, consider 0.

Because the horizontal distance of point C extends 1 unit to the left than that of point A, and point D necessarily extends 1 unit to the right than that of point B, CD is 2 longer than AB, so CD = (1-a)/a+2 = (1+a)/a.

Let the heights of triangle PCD and triangle PAB be h 1, H2 respectively;

h 1+H2 = 1； - 2

Because the triangular PCD is similar to the triangular PAB (which is obviously available)

Therefore, the ratio of h 1 and h2=CD to AB means H1:H2 = (1-a)/A: (1+a)/A-3.

According to 23, we can get h1= (1+a)/2h2 = (1-a)/2.

So s 1-s2= area of triangle PCD minus area of triangle PAB.

=0.5*h 1*| CD| -0.5*h2* |AB|

=0.5[(a+ 1)^2/2a-( 1-a)^2/2a)

=4a/4a

= 1

(Note: This kind of topic is really time-consuming. Asking such questions is not recommended. Sadly, I can't even get on the map (I didn't reach Baidu level 2). I hope it will help you solve the problem! )