(a) Draw ALL the diagonals. Call $\theta_{XY}$ the angle inscribed in arc $\overarc{\(XY\)}$. For example, we get$$\angle A = \theta_{BC} + \theta_{CD} + \theta_{DE} + \theta_{EF}.$$If we do the same thing for $\angle B$, $\angle C$, $\angle D$,$\angle E$ and $\angle F$, we find that $\angle A = \theta_{BC} + \theta_{CD} + \theta_{DE} + \theta_{EF}.$ $\angle B = \theta_{CD} + \theta_{DE} + \theta_{EF}+\theta_{FA} .$ $\angle C = \theta_{DE} + \theta_{EF}+\theta_{FA}+\theta_{AB} .$ $\angle D= \theta_{EF}+\theta_{FA}+\theta_{AB}+\theta_{BC} .$ $\angle E= \theta_{FA}+\theta_{AB}+\theta_{BC} + \theta_{CD} .$ $\angle F= \theta_{AB}+\theta_{BC} + \theta_{CD}+\theta_{DE} .$ Now, we can rewrite the equation $\angle A + \angle C + \angle E = \angle B + \angle D + \angle F$ as $$\theta_{BC} + \theta_{CD} + \theta_{DE} + \theta_{EF}+ \theta_{DE} + \theta_{EF}+\theta_{FA}+\theta_{AB} +\theta_{FA}+\theta_{AB}+\theta_{BC} + \theta_{CD}= \theta_{CD} + \theta_{DE} + \theta_{EF}+\theta_{FA}+\theta_{EF}+\theta_{FA}+\theta_{AB}+\theta_{BC}+ \theta_{AB}+\theta_{BC} + \theta_{CD}+\theta_{DE}$$which is true after mass cancellation.

Thanks for part (b). I was trying sooooo hard to show it was true, but should have found a counter example

easier proof for (a). $\angle BAD+\angle BCD=\angle ABC+\angle ADC$; $\angle DAF+\angle DEF=\angle ADE+\angle AFE$. add this, you will get it.

Lol. Looking at @franzliszt solution, I would've gotten a headache just by typing it.