Answer
$m\angle ABO=18^{\circ}$
Work Step by Step
We know that the $m\angle A=36^{\circ}$. If we draw in the radius $\overline{OA}$ we will create an isosceles triangle; $\triangle BOA$, with congruent base angles of $\angle OBA$ and $\angle BAO$. The measure of central angle $BOA$ is equal to the measure of its intercepted arc, which is $144^{\circ}$ (we found that in part b).
$180-144=36\div2=18$
$m\angle ABO=18^{\circ}$