解:(1)∠A=2∠A1 ∵∠ABC与∠ACD的平分线相交于点A1 ∴∠ABA1=∠A1BC,∠ACA1=∠A1CD ∵∠A1=∠A1BC—∠A1CD ∴∠A=∠ACD—∠ABC=2∠A1BC—2∠A1CD=2(∠A1BC—∠A1CD)=2∠A1 (2)∠A5=96°×(1/2)的五次方=96°×1/32=3°