Do y? 0, get (x? 3)2? 1? 0,

The solution is x 1? 3? ，x2？ 3? .

Point a is to the left of point b,

∴A point coordinates (3? , 0), point B coordinates (3? ,0).

(2) D is DG⊥y axis, and the vertical foot is G. 。

Then G(0,? 1)，GD？ 3.

Do x? 0 and then y? The coordinate of point ∴C is (0,).

∴GC(？ 1)? .

Let the axis of symmetry intersect the x axis at point m.

∵OE⊥CD，

∴∠GCD？ ∠COH？ 90? .

Meng? ∠COH？ 90? ,

∴∠MOE？ ∠GCD。

CGD again? ∠OMN？ 90? ,

∴△DCG∽△EOM.？

∴ .

∴EM？ 2, that is, the coordinate of point E is (3,2), ED? 3.

From Pythagorean Theorem, AE2? 6、AD2？ 3,

∴AE2？ AD2？ 6? 3? 9? ED2。

∴△AED is a right triangle, that is, ∠DAE? 90? .

Let AE pass through the CD at point F.

∴∠ADC？ ∠AFD？ 90? .

AEO again? ∠HFE？ 90? ,

∴∠AFD？ HFE，

∴∠AEO？ ∠ADC。

(3) If the radius of ⊙E is 1, we can get PQ2? EP2？ 1.

In order to minimize the tangent length PQ, it is only necessary to minimize the EP length, that is, EP2.

Let p coordinate be (x, y), and we can get EP2 from Pythagorean theorem. (x？ 3)2? (y？ 2)2.

∵y？ (x？ 3)2? 1,

∴(x？ 3)2? 2y？ 2.

∴EP2？ 2y？ 2? y2？ 4y？ four

? (y？ 1)2? 5.

When will you? At 1, the minimum value of EP2 is 5.

Play y? 1 becomes y? (x？ 3)2? 1, get (x? 3)2? 1? 1,

The solution is x 1? 1，x2？ 5.

Point p is on the parabola on the right side of the symmetry axis,

∴x 1？ 1 give up.

The coordinate of point p is (5, 1).

At this time, the coordinate of point Q is (3, 1) or ().