A drawing can help.
1. The sum of angles of ΔABC is 180°, so ∠B is ...
... ∠B = 180° - ∠A - ∠C = 180° -82° -68° = 30°
ΔABE is isosceles, so ∠EAB = ∠B = 30°.
The sum of angles CAE and EAB will be the measure of ∠A, so we have ...
... ∠CAE + ∠EAB = ∠A
... ∠CAE + 30° = 82° . . . fill in the known values
... ∠CAE = 52° . . . . . . . . subtract 30°
2. ∠CPQ is an external angle to the internal angle CPA of ΔCPA. Thus it is the sum of angles PAC and PCA. Since ΔCPA is isosceles, ∠PAC = ∠PCA and each is half of their sum. That is
... ∠PCA = ∠CPQ/2 = 39°
Likewise, ∠CPQ is an external angle to ΔCQB so is the sum of the equal angles QBC and QCB. Then ...
... ∠QCB = ∠CQP/2 = 31°
The sum of angles in ΔCPQ is 180°, so we have ...
... ∠CPQ + ∠CQP + ∠PCQ = 180°
... 78° + 62° + ∠PCQ = 180° . . . fill in the given values
... ∠PCQ = 40° . . . . . . . . . . . . . .subtract 140°
The angle of interest is the sum of ∠ACP, ∠PCQ, and ∠QCB, so is ...
... ∠ACB = 39° + 40° + 31°
... ∠ACB = 110°
