解:(1)能;
(2)① 22.5°,
②∵AA1=A1A2=A2A3=1,A1A2⊥A2A3,
∴A1A3=,AA3=,
又∵A2A3⊥A3A4 ,
∴A1A2∥A3A4,
同理:A3A4∥A5A6,
∴∠A=∠AA2A1=∠AA4A3=∠AA6A5,
∴AA3=A3A4,AA5=A5A6
∴a2=A3A4=AA3=,
a3=AA3+ A3A5=a2+ A3A5,
∵A3A5=a2,
∴a3=A5A6=AA5=;
(3),,;
(4)由题意得:,
∴。
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