1)
To find the coordinates of the point of the parabola that is at minimum distance from the circle, we need to find the intersection point(s) of the parabola and the normal line(s) to the circle at its point(s) of tangency with the parabola.

First, we'll find the equation of the circle in standard form:

x^2 + (y-6)^2 = 1

Expanding and simplifying:

x^2 + y^2 - 12y + 35 = 0

Next, we'll find the equation of the parabola in vertex form:

y^2 = 8x

Dividing both sides by 8:

y^2/8 = x

So the vertex of the parabola is at (0,0), and the focus is at (2,0).

Now we'll find the point(s) of tangency between the circle and the parabola. Since the vertex of the parabola is at the origin, the tangent line(s) to the circle at its point(s) of tangency will pass through the focus of the parabola.

The distance between the center of the circle and the focus of the parabola is:

d = |2-0| = 2

So the equation(s) of the normal line(s) to the circle at its point(s) of tangency with the parabola will be:

x = ±2

Substituting these values of x into the equation of the parabola, we get:

y^2 = 8(±2)

y^2 = ±16

So the two points of tangency are:

(2,4) and (2,-4)

Now we need to find the coordinates of the point(s) on the parabola that are at minimum distance from these points of tangency.

The distance between a point on the parabola (x,y) and a point of tangency (2,±4) is:

d = sqrt((x-2)^2 + (y-4)^2)

We want to minimize this distance, so we'll square it and minimize the resulting expression:

d^2 = (x-2)^2 + (y-4)^2

Substituting y^2 = 8x, we get:

d^2 = (x-2)^2 + (8x-4y+16)^2/16

Taking the partial derivative with respect to x and setting it equal to 0 to find the critical point(s):

d^2_x = 2(x-2) + (8x-4y+16)(8/16) = 0

Simplifying:

5x - 2y + 8 = 0

Substituting y^2 = 8x, we get:

5y^2/8 - 2y + 8 = 0

Multiplying by 8 to eliminate the fraction:

5y^2 - 16y + 64 = 0

Solving for y using the quadratic formula:

y = (16 ± sqrt(16^2 - 4*5*64))/10

y = (16 ± 8sqrt(5))/10

y ≈ 2.