a) Since ABGF is a rhombus, △ABF and △BGF are congruent and isosceles triangles

∠AFB = ∠ABF = ∠BFG = ∠FBG = 27°
∠DBG = 180° – ∠ABF – ∠FBG – ∠CBD
= 180° – 27° – 27° – 25°
= 101°

b) ∠CBD = ∠BDG = 25°
∠BDE = ∠BDG + GDE
= 25° + 83°
= 108°

Since BD ∥ FE,
∠BDE + ∠DEF = 180°
108° + ∠DEF = 180°
∠DEF = 180° – 108°
= 72°