Calculate the interior angle of a regular pentagon.

A regular pentagon has $n=5$ sides. The formula for the interior angle of a regular $n$-sided polygon is $\frac{(n-2) \times 180^\circ}{n}$.

$$\angle EAB = \frac{(5-2) \times 180^\circ}{5} = \frac{3 \times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ$$

a) $\boxed{\text{108}^\circ}$

In a regular pentagon, all sides are equal in length. Therefore, $EA = AB$.
This means $\triangle EAB$ is an isosceles triangle.
The base angles of an isosceles triangle are equal. So, $\angle ABE = \angle AEB$.
The sum of angles in a triangle is $180^\circ$.

$$\angle EAB + \angle ABE + \angle AEB = 180^\circ$$ $$108^\circ + 2 \times \angle ABE = 180^\circ$$ $$2 \times \angle ABE = 180^\circ - 108^\circ$$ $$2 \times \angle ABE = 72^\circ$$ $$\angle ABE = \frac{72^\circ}{2} = 36^\circ$$

Since N lies on EB, $\angle ABN = \angle ABE$.

b)