a.
We have $\hat{B} = 50^\circ + 30^\circ = 80^\circ$.
Suppose that $\hat{A} = \hat{B} = 80^\circ$. Then
$$\hat{A} + \hat{B} + \hat{\Gamma} = 80^\circ + 80^\circ + \hat{\Gamma} > 160^\circ + 55^\circ = 215^\circ,$$
absurd.
If $A\Gamma = \frac{180^\circ - 80^\circ}{2} - 50^\circ$, then $B\hat{\Gamma}\Delta = 55^\circ < \hat{\Gamma} = 50^\circ$, absurd.
Hence we have: $\Delta\hat{\Gamma}A = 80^\circ - 55^\circ = 25^\circ$ (1).

b.
Let $AZ$ be the bisector of the angle $\hat{A}$. Then $AZ$ is height and median of the triangle $AB\Gamma$. Let $AZ$ meet line $BD$ at point $E$. Since in the triangle $B\Gamma\Delta$ we have $\Gamma\hat{B}\Delta < B\hat{\Gamma}\Delta$, it follows that $\Delta\Gamma < \Delta B$. Hence $\Delta$ lies in the semi-plane with respect to $AZ$ containing point $\Gamma$. It means that $E$ lies between points $B$ and $\Delta$.
Since $EB = E\Gamma$, it follows that $E\hat{\Gamma}B = E\hat{B}\Gamma = 30^\circ$ and hence

$$E\hat{\Gamma}\Delta = 55^\circ - 30^\circ = 25^\circ = \Delta\hat{\Gamma}A. \qquad (2)$$
Hence $\Gamma\Delta$ bisects the angle $E\hat{\Gamma}A$ of the triangle $AE\Gamma$. Moreover, for the external angles $\Delta\hat{E}\Gamma$ and $\Delta\hat{E}A$ of the triangles $EB\Gamma$ and $EBA$, respectively, we have: $\Delta\hat{E}\Gamma = 30^\circ + 30^\circ = 60^\circ$ and $\Delta\hat{E}A = E\hat{B}A + \frac{\hat{A}}{2} = 50^\circ + 10^\circ = 60^\circ$.
Hence $\Delta\hat{E}\Gamma = \Delta\hat{E}A = 60^\circ$ and so $E\Delta$ bisects the angle $A\hat{E}\Gamma$ of the triangle $AE\Gamma$.
Hence $\Delta$ is the incenter of the triangle $AE\Gamma$ and $\Delta\hat{A}\Gamma = \frac{E\hat{A}\Gamma}{2} = \frac{10^\circ}{2} = 5^\circ$.