(1) First, calculate the length of DO according to the properties of the rectangle, and then get the value of t;

(2) If △PQB is to be a right triangle, it is obviously only ∠ PQB = 90 or ∠ PBQ = 90, and then Pb 2 = (6-t) 2+(2-t) 2 and QB 2 = (6-2t) are obtained by Pythagorean theorem.

(3) There is such a t value. If △PQB rotates around a certain point 180, and the corresponding three vertices just fall on a parabola, then the center of rotation is the midpoint of PQ. At this time, the quadrilateral PBQB' is a parallelogram, and the value of t can be obtained according to the properties and symmetry of the parallelogram.

Answer:

Solution: (1)∵ Quadrilateral OABC is a rectangle,

∴∠AOC=∠OAB=90，

∫OD splitting ∠AOC,

∴∠AOD=∠DOQ=45，

∴, ∠ ADO = 45 in Rt△AOD,

∴AO=AD=2,OD=2√2，

∴t=2√2/2=2；

(2) To make △PQB a right triangle, obviously only ∠ PQB = 90 or ∠ PBQ = 90.

As shown in figure 1, let PG⊥OC be at G point and at Rt△POG.

∫∠poq = 45 ,∴∠opg=45，

∵OP=√2t,∴OG=PG=t，

Point P(t, t)

Q (2t，0)，B(6，2)，

According to Pythagorean theorem, we can get: Pb 2 = (6-t) 2+(2-t) 2, QB 2 = (6-2t) 2+2 2, PQ 2 = (2t-t) 2+t2 = 2t 2,

① if ∠ pqb = 90, then PQ2+BQ2 = Pb 2.

Namely: 2t 2+[(6-2t) 2+2 2] = (6-t) 2+(2-t) 2,

Finishing: 4t 2-8t = 0,

Solution: t 1=0 (truncation), t2=2,

∴t=2，

② If ∠ pbq = 90, there is Pb 2+QB 2 = PQ 2,

∴[(6-t)^2+(2-t)^2]+[(6-2t)^2+2^2]=2t^2，

Finishing: t 2- 10t+20 = 0,

Solution: t = 5 √ 5.

∴ When t=2 or t=5+√5 or t=5-√5, △PQB is a right triangle.

Solution 2: ① As shown in Figure 2, when ∠PQB = 90°,

∴bq∠od∴∠bqc =∠poq = 45, Yi Zhi ∠ OPQ = 90.

Available QC=BC=2, ∴OQ=4,

∴2t=4，

∴t=2，

② As shown in Figure 3, when ∠ pbq = 90, if point Q is on OC,

Let the PN⊥x axis intersect at point n and AB at point m,

It is easy to prove that ∠PBM=∠CBQ,

∴△PMB∽△QCB

∴PM/MB=QC/BC，

∴CB？ PM=QC？ MB，

∴2(t-2)=(2t-6)(6-t)，

T 2- 10t+20 = 0，

Solution: t = 5 √ 5,

∴t=5-√5； ?

③ As shown in Figure 3, when ∠ pbq = 90, if point Q is on the extension line of OC,

Let the PN⊥x axis be at point n and intersect with the AB extension line at point m,

It is easy to prove that BPM = MBQ = BQC,

∴△PMB∽△QCB，

∴PM/MB=QC/BC，

∴CB？ PM=QC？ MB，

∴2(t-2)=(2t-6)(t-6)，

T 2- 10t+20 = 0，

Solution: t = 5 √ 5,

∴t=5+√5； ?

(3) There is such a t value for the following reasons:

Rotate △PQB around a point 180, and the corresponding three vertices just fall on a parabola.

Then the center of rotation is the midpoint of PQ, and the quadrilateral PBQB' is the parallelogram.

PO = PQ, from P(t, t) and Q(2t, 0), the coordinate of the rotation center can be expressed as (3/2t, 1/2t).

∫ The coordinate of point B is (6,2), and the coordinate of point B is (3t-6, t-2).

Substituting y =- 1/t (x-t) 2+t, we get: 2t 2- 13t+ 18 = 0,

Solution: t 1=9/2, t2 = 2.