(2) When △FMN is a right triangle, △QWP is also a right triangle. When MF⊥FN, it is proved that △DFM∽△GFN has DF: FG = DM: GN, 4-x=2x, and the value of x at this time is obtained. When MG⊥FN, point M coincides with point A.

(3) When points F, M and N are on the same straight line, MN is the shortest. Let the elapsed time be x, the length of AM be (4-x) and the length of AN be (6-x), then the answer can be obtained from △MAN∽△MBF. Solution: (1) ∵.

∴∠QPW=∠PWF,∠PWF=∠MNF，

∴∠QPW=∠MNF.

Similarly ∠PQW=∠NFM,

∴△fmn∽△qwp；

(2) Because △FMN∽△QWP, when △FMN is a right triangle, △QWP is also a right triangle.

Let FG⊥AB, then the quadrilateral FCBG is a square, GB=CF=CD-DF=4, GN=GB-BN=4-x, DM=x,

① When MF⊥FN,

∠∠DFM+∠MFG =∠MFG+∠GFN = 90，

∴∠DFM=∠GFN.

∠∠D =∠FGN = 90，

∴△DFM∽△GFN，

∴DF:FG=DM:GN=2:4= 1:2，

∴GN=2DM，

∴4-x=2x，

∴x= 43；

② When mg ⊥ fn, point M coincides with point A, and point N coincides with point G,

∴x=AD=GB=4.

∴ When x=4 or 43, △QWP is a right triangle; When 0 < x < 43 and 43 < x < 4, △QWP is not a right triangle.

(3) When points M, N and F are on the same straight line, MN is the shortest.

Let the elapsed time be x and the length of AM be (x-4).

The length of AN is (6-x),

The length of MN is the hypotenuse of a right triangle,

MN2 = AM2+AN2 =(x-4)2+(6-x)2 = 2 x2-20x+52，

When x=5, the minimum value of MN2 is 2,

So the shortest hypotenuse of MN is √ 2.