1. angle BAC=1/2 angle BOC, and BOC= 360-(110+120)=130, BAC=130/2=$\boxed{65^\circ}$
2.
If NM is the median of trapezoid ABCD, then that means NM is the average of the two heights, which in this case are 5 and 17. so that means NM= $\frac{5+17}2=11$
now to find the median of trapezoid ABMN, we have to do the same thing, take the average of the two heights, which in this case are 5 and 11. This means that PQ=$\frac{5+11}2=\boxed{8}$
3.
we can first start off by figuring out how many degrees are in each interior angle of a hexagon: (6-2)*180$^\circ$=720$^\circ$ $\frac{720^\circ}{6}=120^\circ$ and we know that the triangle with B and C as the two base vertices is an isoscolese, because the square and the hexagon have equal side lengths, so angle B = angle C. so we have
$\frac{180-(120-90)}2$= angle C = $75^\circ$
and because we want to find angle ABC, we have to subtract 75$^\circ$ from 120 to get angle ABC = 120-75 = $\boxed{45^\circ}$
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