Here is an excellent solution from Andrei Lazanu, age 14, School No. 205, Bucharest, Romania.

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 $O_{1} {AO}_{2}$ and $O_{1} {BO}_{2}$ 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 $O_{1} {AO}_{2}$ and $O_{1} {BO}_{2}$ are equal to say $θ$ . A tangent to a circle and the radius to the point of contact are perpendicular so $∠ {MAO}_{2} = ∠ {NAO}_{1} = ∠ {MBO}_{2} = ∠ {NBO}_{1} = 90^{o}$ . Hence $∠ O_{1} AM = ∠ O_{2} AN = ∠ O_{1} BM = ∠ = ∠ O_{2} BAN = θ - 90^{o}$ . So, $∠ MAN = ∠ MBN$ .

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 $T_{1} T_{1}'$ , $T_{2} T_{2}'$ , $T_{3} T_{3}'$ 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.
Andrei's Solution
From the previous point, I see that angle $α$ has the same measure with angle $T_{2} {DT}_{1}'$ and to the angle $T_{1} {DT}_{2}'$ (the last two being vertically opposite angles).

In the same manner I found: $β = ∠ T_{1}' {DT}_{3}' = ∠ T_{1} {DT}_{3}$ and $γ = ∠ T_{3} {DT}_{2} = ∠ T_{3}' DT 2'$ .

This way, it is easy to see that the three angles add up to $180^{o}$ : $α + β + γ = T_{2} {DT}_{1}' + T_{1}' {DT}_{3}' + T_{3}' {DT}_{2}' = T_{2} {DT}_{2}' = 180^{o}$ .

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

Andrei's Solution
I find $α = ∠ T_{2} {DT}_{1}'$ , $β = ∠ T_{2}' {DT}_{3}' = ∠ T_{2} {DT}_{3}$ and $γ ∠ T_{3} {DT}_{1} = ∠ T_{3}' {DT}_{1}'$ .

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