Part (b) follows directly from part (a). We have $\dfrac{AB}{DC}=\dfrac{1}{2}$. Using the expressions you got for $AB$ and $CD$, putting $\beta=2\alpha$, and simplifying a bit, we find that $\frac{1}{2}=\frac{AB}{CD}=\frac{\sin 4\alpha}{\sin 2\alpha}$ (there is a fair amount of cancellation.)
Now use the double-angle identity $\sin 4\alpha=2(\sin 2\alpha)(\cos 2\alpha)$. We find that $\cos 2\alpha =\frac{1}{4}.$ The rest is a job for the calculator, use the $\cos^{-1}$ button.
If we feel like it we can get an explicit expression for $\cos \alpha$ or $\sin \alpha$, from the fact that $\cos 2\alpha=2\cos^2\alpha-1=1-2\sin^2\alpha$, but that is not necessary.
As to part (c), one cannot inscribe a circle in any trapezoid that satisfies the conditions of (b). First draw an isosceles trapezoid with an inscribed circle. Suppose that the two tangents to a circle from an external point $P$, meet the circle at $M$ and $N$. Then $PM=PN$. So in any isosceles trapezoid with an inscribed circle, each "slant" side has length equal to half the sum of the two parallel sides.
In our case, we can take $AB=2$ and $CD=4$. So if our trapezoid had an inscribed circle, then each slant side would have length $3$. But then the height of the trapezoid would be $\sqrt{3^2-1^2}$, that is, $2\sqrt{2}$. That makes $\tan\alpha =\frac{2\sqrt{2}}{3}$. But then $\angle \alpha\approx 43.3$ degrees, which is not what we found in part (b).