解：?$(1)$?因?yàn)?$OA = OB，$??$∠ABO = 30°，$?
所以?$∠A=∠ABO = 30°。$?
?$ $?因?yàn)?$∠A+∠ABO+∠AOB = 180°，$?
所以?$∠AOB=180°-∠A - ∠ABO=180°-30°-30°=120°。$?
?$ $?因?yàn)橹本€?$MN$?是?$\odot O$?的切線，
所以?$EC\perp MN，$?即?$∠ECM = 90°。$?
?$ $?又因?yàn)?$AB// MN，$?
所以?$∠CDB=∠ECM = 90°。$?
?$ $?在?$Rt\triangle BDO$?中，?$∠BOE = 90°-∠ABO=90°-30°=60°。$?
?$ $?因?yàn)?$\overset {\frown }{BE}=\overset {\frown }{BE}，$?
所以?$∠BCE=\frac {1}{2}∠BOE = 30°。$?
?$(2)$?連接?$OC，$?則?$OC = OA = 3。$?
?$ $?因?yàn)橹本€?$MN$?是?$\odot O$?的切線，
所以?$OC\perp MN。$?
?$ $?又因?yàn)?$OB// MN，$?
所以?$OC\perp OB，$?即?$∠COB = 90°。$?
?$ $?因?yàn)?$CG\perp AB，$?
所以?$∠FGB = 90°。$?
?$ $?因?yàn)?$\triangle COF $?與?$\triangle FGB$?的內(nèi)角和都為?$180°，$??$∠OFC=∠BFG，$?
所以?$∠OCF=∠ABO = 30°。$?
?$ $?在?$Rt\triangle COF_{中}，$?設(shè)?$OF = x，$?則?$CF = 2x。$?
?$ $?由勾股定理?$OC^2+OF^2=CF^2，$?即?$3^2+x^2=(2x)^2，$?
?$ 9 + x^2=4x^2，$?
?$ 3x^2=9，$?
?$ x^2=3，$?
?$ $?解得?$x = \sqrt {3}($?負(fù)值舍去?$)，$?
所以線段?$OF $?的長(zhǎng)為?$\sqrt {3}。$?
