a.
From the right angled triangle $B\Gamma\Delta$ we have: $\hat{\Gamma}_1 = 90^\circ - \hat{B}$.
The line joining the centers of the circles $c_1$ and $c_2$ is the perpendicular bisector of their common chord $\Delta E$.
If $T$ is the point of intersection of the lines $B\Gamma$ and $\Delta E$, then:
$$\hat{\Gamma}_1 + \hat{\Delta}_1 + \hat{\Delta}_2 = 90^\circ, \text{ that is } \hat{\Delta}_1 + \hat{\Delta}_2 = \hat{B} = \hat{\Gamma}. \quad (1)$$
From the isosceles triangle $\Gamma\Delta Z$ we have: $\hat{\Delta}_1 = \hat{Z}_1$ and $\hat{\Gamma}_2 = 2\hat{\Delta}_1 = 90^\circ - \hat{A}$, and hence:
$$\hat{\Delta}_1 = 45^\circ - \frac{\hat{A}}{2}. \quad (2)$$
From the isosceles triangle $AB\Gamma$ we have: $\hat{B} = 90^\circ - \frac{\hat{A}}{2}$ \quad (3)
From (1), (2), (3) we get:
$$\hat{\Delta}_1 + \hat{\Delta}_2 = \hat{B} = \hat{\Gamma} \Leftrightarrow \hat{\Delta}_2 = \hat{B} - \hat{\Delta}_1 \stackrel{(2),(3)}{\Leftrightarrow} \hat{\Delta}_2 = 90^\circ - \frac{\hat{A}}{2} - \left( 45^\circ - \frac{\hat{A}}{2} \right) = 45^\circ.$$

b.
The angle $\hat{\Delta}_1$ is created from the chord $\Delta M$ and the tangent $\Delta\Gamma$ of the circle $c_1$, and hence $\Delta\hat{E}M = \hat{\Delta}_1$. Also we have $\Delta\hat{E}K = \Delta\hat{Z}K = \hat{Z}_1$.
Since $\hat{\Delta}_1 = \hat{Z}_1$, we get $\Delta\hat{E}M = \Delta\hat{E}K$, and the points $E$, $M$, $K$ are collinear.

c.
We will prove that $\Delta\hat{B}M = 90^\circ - \hat{A}$.
We have $\Delta\hat{B}M = 2\Delta\hat{E}M = 2\hat{\Lambda}_1 = \hat{\Gamma}_2 = 90^\circ - \hat{A}$, and hence $BM \perp A\Gamma$. Moreover $E\hat{K}Z = E\hat{\Delta}Z = \hat{\Lambda}_2 = 45^\circ$, and so the orthogonal triangle $EZK$ isosceles. Then its median $E\Gamma$ is also an altitude, that is $E\Gamma \perp KZ$. Hence $BM \parallel E\Gamma$ (both are perpendicular to the line $AZ$).

Figure 2