As shown in the figure, let QM vertical AB intersect with O in Q', connect PQ' with AB in G, connect GQ, let Q'N vertical PF, let DH vertical AB, and the vertical feet are N and H respectively.

Solution: (1) Because PD: DC = 2: 3, FC=DC=3, so PD=2, PC=5, OC=5, so PC=OC=5, and we know that EC is the bisector of angle PCO, so angle PCE= angle OCE, we can get the triangle PCE congruent triangles OCE. Therefore, the angle PDO=90 degrees, so OD is perpendicular to PC, and PC is tangent to circle O, and the square of DC =CAXCB can be obtained. If the radius of circle O is r and the square of 3 =(5+R)X(5-R), the solution is R=4.

(2) Because OQ=OD and DOQ=90 degrees, MOQ= ODH and QMO= OHD=90 degrees, so the triangle QMO is congruent triangles OHD and ODXDC=OCXDH, so 4X3=5XDH, DH= 12/5 and OM = DH = 650.

OF=5-3=2, MF=Q'N=22/5, FN=OM=QM= the square of 4 under the radical number -( 12/5) 16/5,

So PG+QG=PG+Q'G=PQ'= the square under the root number (16/5+4)+the square under the root number (12/5+2) = 445.

Therefore, the minimum value of PG+QG is 445 under 2/5 roots.

(3) When Q and A coincide, the maximum area of triangle PQE is1/2xpXaf =1/2x5/2x6 =15/2.

Pc: fc = PE: ef can be obtained from the bisector theorem of inner angle, so it is 5: 3 = PE: (4-PE) and PE=5/2.