解：如图，∵四边形ABCD是平行四边形，∠ADC=125°，∴AB∥CD，∠ABC=∠ADC=125°，∴∠ADC+∠DAB=180°，则∠DAB=180°-125°=55°．又∵∠CAD=21°，∴∠CAB=∠DAB-∠CAD=55°-21°=34°．综上所述，∠ABC、∠CAB的度数分别是125°、34°．
