Point p, point q and point w are the midpoint of FM, MN and NF respectively.

∴pq= 12fn,qw= 12fm,pw= 12mn.

∴FNPQ=FMQW=MNPW=2.

∴△FMN∽△QWP.

(2) solution: ①∵ quadrilateral ABCD is rectangular,

∴∠DAB=∠B=∠C=∠D=∠90，AD=BC=4cm，DC=AB=6cm。

From the title, it can be seen that DM = BN = 1× t = t (cm).

So AM=.4? t .，AN=6-t。

∴MN2=AM2+AN2

=(4-t)2+(6-t)2

=2t2-20t+52

=2(t-5)2+2。

∵2>0,

When t=5, MN2 is the smallest, and the minimum value is 2.

When t=5, MN takes the minimum value and the minimum value is 2.

②∫△FMN∽△QWP，

∴∠PQW=∠NFM,∠QWP=∠FMN,∠WPQ=∠MNF.

1. When ∠ PQW = 90, ∠ NFM = 90.

The intersection n is NE⊥DC and the vertical leg is e, as shown in Figure 2.

∠∠MDF =∠MFN =∠FEN = 90，

∴∠DFM=90 -∠EFN=∠ENF。

∴△MDF∽△FEN.

∴DMEF=DFEN.

DM = t，DF=2，EF=CF-CE=6-2-t=4-t，EN=BC=4，

∴t4？ t=24。

Solution: t = 43.

It is verified that t=43 is the solution of the equation, which accords with the meaning of the question.

Ⅱ. When ∠ PWQ = 90, ∠ NMF = 90.

Similarly: △ MDF ∽△ Nam.

There is DFAM=DMAN ..

DF = 2，DM=t，AM=4-t，AN=6-t，

∴24? t=t6？ t.

Finishing: T2-6t+ 12 = 0.

∵(-6)2-4× 1× 12=- 12