【解答】证明:(1)连接OT;∵OT=OA,∴∠ATO=∠OAT,又∵∠TAC=∠BAT,∴∠ATO=∠TAC,∴OT∥AC;∵AC⊥PQ,∴OT⊥PQ,∴PQ是⊙O的切线.(2)解:过点O作OM⊥AC于M,则AM=MD;又∠OTC=∠ACT=∠OMC=90°,∴四边形OTCM为矩形,∴OM=TC=3,∴在Rt△AOM中,AM=OA2-OM2=4-3=1,∴弦AD的长为2.