(1) connects OC, and the tangent of ∠ ABC+∠ BAC = 90 and CM⊙O gives ∠ ACM+∠ ACO = 90, and then draws a conclusion with ∠BAC=∠AOC.

(2) Connect OC to obtain that △AEC is a right triangle and the diameter of the circumscribed circle of △AEC is AC. Get AC with △ABC∽△CDE.

Answer: (1) Proof: As shown in the figure, connect OC,

∵AB is the diameter⊙ O,

∴∠ACB=90，

∴∠ABC+∠BAC=90，

And ∵CM is the tangent of ⊙O,

∴OC⊥CM，

∴∠ACM+∠ACO=90，

CO = AO，

∴∠BAC=∠ACO，

∴∠acm=∠abc；

(2) solution: BC = CD,

∴OC∥AD，

∵OC⊥CE，

∴AD⊥CE，

∴△AEC is a right triangle,

The diameter of the circumscribed circle of AEC is AC,

∠∠ABC+∠BAC = 90，∠ ACM+∠ ECD = 90，

∴△ABC∽△CDE，

∴AB/CD=BC/ED，

⊙O has a radius of 3,

∴AB=6，

∴6/CD=BC/2，

∴BC^2= 12，

∴BC=2√3，

∴AC=√(36？ 12)=2√6,

∴△△△ The radius of the circumscribed circle of AEC is ∴ 6.