(1) Given that the smaller inner angle of the diamond is 60 degrees, it is easy to get BD=AD, angle BDF= angle DAE, and DF=AE, and the conclusion is valid.

(2) (Make a triangle congruent with DGC as an auxiliary line by rotating) Take BC as one side, and make an angle BCM outside the rhombus, so that the angle BCM= angle DCG, CM exceeds GB and extends to point M. ..

Angle DGC= 180 degrees-Angle GDC- Angle DCG= 180 degrees -( 120 degrees-Angle ADE)- Angle DCG=60 degrees+Angle ADE- Angle DCG.

Angle CMB= Angle GBC- Angle BCM=60 degrees+Angle DBF- Angle BCM.

By (1), angle ADE= angle DBF, and by, angle BCM= angle DCG.

So triangle CBM equals triangle CDG, then angle GCM= angle DCB=60 degrees, and triangle CGM is an equilateral triangle, then the area of quadrilateral CBGD = triangle CGM= three times the square of the side length (i.e. CG). The conclusion is valid.

(3) The intersection point E makes the parallel DA of EN intersect FB at point N. From AF: FD = 2: 1, we can easily get EB: AE = 2: 1 = BN: FN, so BN=2FN, and the triangle BNE is similar to the triangle BFA. When EN: AF = 2: 3, then EN: DF.

Because triangle GEN is similar to triangle GDF, NG: FG = EN: DF = 4: 3, so FG=(3/7)FN, NG=(4/7)FN, and the conclusion is valid through calculation.