Answer: (1) Proof: ∵b=2a, and point m is the midpoint of AD.
∴AB=AM=MD=DC=a，
∫ In rectangular ABCD ∠ A = ∠ D = 90,
∴∠AMB=∠DMC=45，
∴∠BMC=90。
(2) solution: existence,
Reason: If BMC = 90,
Then ∠ AMB = ∠ DMC = 90,
And ≈AMB+∠ABM = 90,
∴∠ABM=∠DMC，
∠∠A =∠D = 90，
∴△ABM∽△DMC，
∴=,
Let AM=x, then =,
Ordered: x2-bx+a2 = 0,
∫b > 2a，a>0，b>0，
∴△=b2﹣4a2>0，
The equation has two unequal real roots, both of which are greater than zero, which accords with the meaning of the question.
When b > 2a, there exists BMC = 90,
(3) Solution: Not valid.
Reason: If BMC = 90,
According to (2), x2-bx+a2 = 0,
∫b < 2a，a>0，b>0，
∴△=b2﹣4a2