Question (i):
In ABOC, ∠OCA = 90° (tangent and the radius meet at perpendicular angle)
∠ABO = 90° (same as above)
∵ ∠OCA + ∠ABO = 180°, ABOC is a cyclic quadrilateral
In AOGC, ∠OCA = 90° (proven)
∠OGA is also 90° (line from centre bisects the chord of a circle at a right angle)
∵∠OCA = ∠OGA, AOGC is a cyclic quadrilateral (equal angles standing on the same arc on a cyclic quad)
Question (ii):
Since ABOC is a cyclic quad, ∠OGF = ∠OAC (exterior angle of a cyclic quadrilateral is equal to the interior opposite angle)
Question (iii):
Let ∠OGF = α
From (ii), ∠OAC = α
AC = AB (tangents meeting at an external point are equal)
AO is common
OC = OB (radii)
∠BAO = α (corresponding angles in congruent triangles)
∠BOC = π - 2α (supplementary angles; opposite angles of a cyclic quad are supplementary)
It should be pretty simple after this, I'll finish it in a little bit.