Green arc : Red arc = 7 : 11

Since $\stackrel{\frown}{ABM}+\stackrel{\frown}{ACM}=360^{\circ}$ use the equation $7x+11x=360^{\circ}$ to get $x=20^{\circ}$ and so $\stackrel{\frown}{ABM}=7x=7\cdot 20^{\circ}=140^{\circ}$

$\stackrel{\frown}{AB} =2\times\angle C = 72^{\circ}$

$\stackrel{\frown}{BM}=140^{\circ}-72^{\circ}=68^{\circ}$

$\angle BAM = \frac{1}{2}\cdot 68^{\circ}=34^{\circ}$ , so $\angle BAC = 68^{\circ}$

$\angle B= 180^{\circ}=68^{\circ}-36^{\circ}=\boxed{76}^{\circ}$

Let $O$ be the circumcenter of $\triangle ABC$ , and $M$ be the intersection point of circumcircle and bisector of $\angle BAC$ . Ratio of lengths of minor and major arc of $AM$ is $7: 11$ , so if $R$ is the circumradius, we have:

$\dfrac{\dfrac{\angle AOM}{360°}2 \pi R}{\dfrac{360° - \angle AOM}{360°}2 \pi R} = \dfrac{\angle AOM}{360° - \angle AOM} = \dfrac{7}{11} \Rightarrow \angle AOM = 140°$

$\angle ACM$ is inscribed, so: $\angle ACM = \dfrac{1}{2}\angle AOM = 70°$ . Also, quadrilateral $ABMC$ is cyclic, so: $\angle BCM = \angle BAM = \dfrac{1}{2} \angle BAC$ .

Now we have that $\angle ACM = \angle BCA + \angle BCM = 36° + \dfrac{1}{2} \angle BAC = 70° \Rightarrow \angle BAC = 68°$ .

Finally, $\angle ABC = 180° - \angle BAC - \angle ACB = \boxed{76°}$