I divided the solution into two parts: the demonstration of the congruence of angles, and the demonstration that the three angles add up to ${180}^o$.

1. Congruence of angles in a lune

Let $O_1$ and $O_2$ be the centres of two circles, $A$ and $B$ the intersection points and $M$ and $N$ the intersections of the tangents to these two circles. I observed that triangles $O_1{\mathrm{AO}}_2$ and $O_1{\mathrm{BO}}_2$ are congruent, because they have a common side $O_1{O}_2$, and the other two sides are radii in the two circles. So, angle $O_1{\mathrm{AO}}_2$ and $O_1{\mathrm{BO}}_2$ are equal to say $\theta$. A tangent to a circle and the radius to the point of contact are perpendicular so $\angle {\mathrm{MAO}}_2=\angle {\mathrm{NAO}}_1=\angle {\mathrm{MBO}}_2=\angle {\mathrm{NBO}}_1={90}^o$. Hence $\angle O_1\mathrm{AM}=\angle O_2\mathrm{AN}=\angle O_1\mathrm{BM}=\angle =\angle O_2\mathrm{BAN}=\theta -{90}^o$. So, $\angle \mathrm{MAN}=\angle \mathrm{MBN}$.

2. Let $O_1$, $O_2$ and $O_3$ be the centres of the three circles respectively, D the point common to all three circles, and $T_1{T}_1\text{'}$, $T_2{T}_2\text{'}$, $T_3{T}_3\text{'}$ the three tangents. In my figure the radius and the tangent to circle 1 are in red, to circle 2 in blue and to circle 3 in green respectively.

From the previous point, I see that angle $\alpha$ has the same measure with angle $T_2{\mathrm{DT}}_1\text{'}$ and to the angle $T_1{\mathrm{DT}}_2\text{'}$ (the last two being vertically opposite angles).

In the same manner I found: $\beta =\angle T_1\text{'}{\mathrm{DT}}_3\text{'}=\angle T_1{\mathrm{DT}}_3$ and $\gamma =\angle T_3{\mathrm{DT}}_2=\angle T_3\text{'}\mathrm{DT}2\text{'}$.

This way, it is easy to see that the three angles add up to ${180}^o$: $\alpha +\beta +\gamma =T_2{\mathrm{DT}}_1\text{'}+T_1\text{'}{\mathrm{DT}}_3\text{'}+T_3\text{'}{\mathrm{DT}}_2\text{'}=T_2{\mathrm{DT}}_2\text{'}={180}^o$.

For the second case, I follow exactly the same steps.

I find $\alpha =\angle T_2{\mathrm{DT}}_1\text{'}$, $\beta =\angle T_2\text{'}{\mathrm{DT}}_3\text{'}=\angle T_2{\mathrm{DT}}_3$ and $\gamma \angle T_3{\mathrm{DT}}_1=\angle T_3\text{'}{\mathrm{DT}}_1\text{'}$.

The result is the same, i.e. that in this case the three angles add up to ${180}^o$ again.