Maths problem: how to do (a), thanks?

(a)
Let n be the number of sides of the polygon.
Then angle sum of interior angles is (n - 2) × 180°.

It is given that the largest interior angle is T(n) = 132°.
Therefore, the smallest interior angle is T(1) = T(n) - (n-1)(12°) = 132° - (n-1)(12°).

Equating the total sum of n interior angles is
[T(1) + T(n)] × n / 2 = (n - 2) × 180°
[264° - (n-1)(12°)] × n / 2 = (n - 2) × 180°
(264 - 12n + 12)(n) = (n - 2)(360)
264n - 12n² + 12n = 360n - 720
12n² + 84n - 720 = 0
n² + 7n - 60 = 0
(n - 5)(n + 12) = 0
n = 5  or  n = -12 (rejected).
That is, the polygon is a pentagon with 5 sides.

(b)
The smallest interior angle is 132° - 4(12°) = 84°.
Therefore, the second smallest interior angle is 84° + 12° = 96°.