Maths problem: how to do , thanks?

(a)
According to the question, ΔABC is an equilateral triangle.
(I think this is the part some candidates omitted?)
Therefore, AB = BC = CA and ∠ABC = ∠BCA = ∠CAB = 60°.
That means,
AB = CA (equilateral Δ)
∠ABD = ∠CAE = 60° (equilateral Δ)
AE = BD (given)
ΔABD ≅ ΔCAE (SAS)

(b)
Let ∠ACE = x, then ∠BAD = x (corr. ∠s, ≅Δs)
∠DAC = ∠BAC - ∠BAD = 60° - x
Finally,
∠DOC
= ∠DAC + ∠ACE (ext. ∠ of Δ)
= x + (60° - x)
= 60°