With (1)D(0, √3/3) as the vertex, let the analytical formula of this parabola be y = a (x-0) 2+√ 3/3 (vertex equation).

Because B(- 1/2, √3/2) is on a parabola, it brings √ 3/2 = A (- 1/2) 2+√ 3/3, and the solution is a=2√3/3.

So the analytical formula y = 2 √ 3/3 * x 2+√ 3/3.

(2) Because A(0, √3) and D(0, √3/3), AD=2√3/3.

Because C (1, 0) and D(0, √3/3), CD=2√3/3.

AD=CD, so ∠DAC=∠DCA,

Because △ABC is folded along AC, ∠DCA=∠B'CA, ∠DAC=∠B'CA, and the internal dislocation angles are equal, ao ∠ CB'

And because △ABC is folded along AC, ∠ AB 'c = ∠ ABC = 90, ∠ B 'ca+∠ B 'ac = 90,

And ∠DAC=∠B'CA, so ∠ DAC+∠ b 'ac = 90, that is ∠ b 'ao = 90.

And ∠ AOC = 90, so ∠AOC=∠B'AO, so AB'‖OC.

That is, the quadrilateral AOCB' is a parallelogram, (according to: two groups of parallelograms with parallel opposite sides are parallelograms)

Because AOC = 90, the quadrilateral AOCB' is a rectangle (according to: a parallelogram with right angles is a rectangle).

So the coordinates of point b' are (1, √3) (where it comes from is not specified),

The equation brought into the parabolic analytical formula y = 2 √ 3/3 * x 2+√ 3/3 holds, so b' is on this parabola.

(3) Draw your own pictures first.

A(0, √3), B(- 1/2, √3/2), so the analytical formula of straight line AB is y=√3x+√3 (will you ask? Do not explain)

E On the straight line AB, let the coordinates of point E be (m, √3m+√3), then the slope of AE is √3 (will you ask? Do not explain)

F on a parabola, let the coordinates of point F be (n, 2 √ 3/2 √ 3/3 * n 2+√ 3/3).

D(0, √3/3), so the slope of DF is 2√3/3 * n (will you ask? Do not explain)

Let two slopes be equal, that is, √3=2√3/3 * n, then the solution is n=3/2. At this time, AP √ DF.

And AD‖PF (because of what? Do not explain)

So the quadrilateral PADF is a parallelogram, and the coordinates of point F are (3/2, 1 1√3/6).

So the coordinate of point P is (3/2, 15 √ 3/6) (will you ask? Do not explain)

I don't know what to ask.

I hope the answer will help you.