Circle Geometry, Equilateral Triangles 
I cannot do part iii,) I considered using angle at circumference and angle at centre, for all 3 circles.
 A: As Aretino said this is well known as Napoleon's Triangle, however I've considered to add the solution here:
Observe that the triangles $BPQ$ and $XBC$ are due to the spiral symilarity of center $B$. Since the Sprial symmilarity carries $X\to C$ and $P\to Q$ because the triangles $BXP$ and $BQC$ are symmilar so the spiral symmilarity also carries $X\to P$ and $C\to Q$. 
Proceed the same way to the triangles $APR$ and $AXC$. This will give you the angle of $\angle RPQ=60^o$ proceeding the same way on the last triangle and you will get what you wanted.
Second proof using what known:
The first part (I) allows you to see that $X,O,C$ are colineal so are $YOA$. The second one allows you to see that $(ZACO)$ and the other two circles are concurrent at $O$.
First observe that traingles $BAY$ and $BCX$ are symmiliar (angle chasing using the colineality).
Now observe that $BPQ$ is also symmilar (angle chasing see that $PQ$ is perpendicular to $BO$).
Using this on the other triangles you can evaluate the angles of the triangle $PQR$ and so see that they are $60^o$ 
A: Let PQ intersect BO at M. What can you say about the size of $\angle BMP$?
Construct N the same way as above from RQ and OC.
Is MONQ cyclic?
If yes, from $\angle MON$, deduce that $\angle PQR = 60^0$.
