解:(1)∵∠AOE+∠DOE=90°;

∠DBA+∠DOE=90°.

∴∠AOE=∠DBA.(同角的余角相等)

又∠OAE=∠BAD.(已知)

∴∠AOE+∠OAE=∠DBA+∠BAD.

即∠OED=∠ODE,得OD=OE=3;

作EH垂直OA于H,HE的延长线交AB于P.

∵∠APE=∠AOE(均为∠CAH的余角);AE=AE;∠PAE=∠OAE.

∴⊿PAE≌⊿OAE(AAS),PE=OE=3.

∵EF∥PB;PE∥BF.

∴四边形BFEP为平行四边形,BF=PE=3.

故OB=OD+DF+BF=3+2+3=8,即点B为(0,8);

(2)∵EF∥AB;OC⊥AB.

∴OE⊥EF,则EF=√(OF²-OE²)=4.

作EM垂直OF于M,由面积关系可知:EF*OE=OF*EM,4*3=5*EM, EM=12/5.

则:CE=HE=OM=√(OE²-EM²)=9/5.

所以,S⊿ACE/S⊿OAE=CE/EO=(9/5)/3=3/5.(同高三角形的面积比等底边之比)