15.如图,在四边形ABCDABCDABCD中,AB//CDAB//CDAB//CD,AB⊥ADAB\bot ADAB⊥AD,CD=1CD=1CD=1,AB=AD=3AB=AD=3AB=AD=3,EEE是BCBCBC的中点,设AB→=a→\overrightarrow{AB}=\overrightarrow{a}AB=a,AD→=b→\overrightarrow{AD}=\overrightarrow{b}AD=b.(1)用a→\overrightarrow{a}a,b→\overrightarrow{b}b表示AE→\overrightarrow{AE}AE;(2)若AF→=2FB→\overrightarrow{AF}=2\overrightarrow{FB}AF=2FB,AEAEAE与CFCFCF交于点OOO,求cos∠AOF\cos \angle AOFcos∠AOF.