a.
The angles $\widehat{AMB}$ and $\widehat{AN\Gamma}$ are right since they see the diameters $AB$ and $A\Gamma$. Therefore the quadrilateral $AMHN$ is cyclic, which is the desired result.

b.
$\hat{T}_1 = \hat{I}_1 \quad (1)$.

The line $E\Delta$ connects the midpoints of $AB$ and $A\Gamma$, so $MN$ is parallel to $B\Gamma$.
Therefore:

$$\hat{r}_1 = \hat{N}_1 \quad (2).$$

From the quadrilateral $AMHN$ we have:
$$\hat{H}_1 = \hat{N}_2 \quad (3).$$
From the relations (1), (2), (3) we have:
$$\hat{T}_1 + \hat{H}_1 = \hat{N}_1 + \hat{N}_2 = A\hat{N}H = 90^\circ.$$

From the quadrilateral $AB\Sigma\Gamma$ we have:
$$A\hat{B}\Gamma = A\hat{\Sigma}\Gamma = \hat{B} \quad \text{and from } E\Delta \parallel B\Gamma \text{ we have: } \hat{E}_1 = \hat{B}.$$
From the equality $A\hat{\Sigma}\Gamma = A\hat{\Sigma}N$ in combination with the previous ones we get that the quadrilateral $AE\Sigma N$ is cyclic.

The quadrilateral $AZ\Sigma N$ is also cyclic, since $A\hat{N}\Sigma = A\hat{Z}\Sigma = 90^\circ$. To this end, we have that $\hat{Z}_1 = \hat{N}_2 = \hat{H}_1$ and so $BH \parallel EZ$. Since $E$ is the midpoint of $AB$, we conclude that $Z$ is the midpoint of $AH$.