BD=√2，？ AC=2, because the rhombic diagonals bisect each other vertically. According to Pythagorean theorem, AB=√6/2, AF = √ (AC 2+CF 2) =? 2√2,
BF=√(BC^2+CF^2)=？ √22/2,
At △ABF, according to the cosine theorem, BF 2ab 2+AF 2-2 * AB * AFCOS < BAF.
cos & ltBAF=√3/3，
AF=AF，AB=AD，BF=DF，
△ABF?△ADF，
BG⊥AF from B, connected to DG,
According to congruent triangles's congruence relation, DG ⊥ AF < BGD is the plane angle of dihedral angle B-AF-D,
Sin< package =√6/3, Sin< bag? =BG/AB，BG= 1，BG=DG= 1，BD=√2，
Triangle BGD is an isosceles right triangle with BGD=90 degrees.
The dihedral angle B-AF-D is 90 degrees.
2. The common part of E-ABCD and F-ABCD is M-ABCD.
MN is the common midline of two triangles AFC and BEC, MN‖CF‖AE,
AE⊥ plane ABCD, CF⊥ plane ABCD,
Therefore, MN⊥ plane ABCD,
MN=CF/2= 1，
S diamond ABCD=BD*AC/2=2*√2/2=√2,
∴V Pyramid M-ABCD=? S diamond ABCD*MN/3=√2* 1/3=√2/3.