解：
(1) 因为∠AOB与∠AOE互为余角，
所以$∠ AOB + ∠ AOE = 90°。$
又因为$∠ AOB = 4∠ AOE，$
所以$4∠ AOE + ∠ AOE = 90°，$
即$5∠ AOE = 90°，$
解得$∠ AOE = 18°，$
所以$∠ AOB = 4×18° = 72°。$
(2) 因为点O在直线AD上，所以$∠ AOD = 180°，$
所以$∠ BOD = 180° - ∠ AOB = 180° - 72° = 108°。$
又因为$∠ COD = 2∠ COB，$且$∠ COD + ∠ COB = ∠ BOD，$
所以$2∠ COB + ∠ COB = 108°，$
即$3∠ COB = 108°，$
解得$∠ COB = 36°。$
