Let $L$ be the midpoint of minor arc $BC$. It is well-known that $LI = LI_A$. Invert about $L$ with radius $LI$ to get $\angle LIM = \angle LNI$ and $\angle LI_AM = \angle LNI_A$, and therefore $\angle IMI_A + \angle INI_A = 180^{\circ}$. As $LN$ is a diameter of $(ABC)$ it follows that $T$ and $N$ are symmetric about $II_A$ and therefore $\angle ITI_A = \angle INI_A$. This combined with the former implies $\angle IMI_A + \angle ITI_A = 180^{\circ}$ which proves the required result.

Let $II_A$ meet $(ABC)$ at $L$. Note that $LI^2=LI_A^2=LB^2=LM \cdot LN$. Thus \[ \angle IMI_A = 180^{\circ}-\angle MIA - \angle II_AM = 180^{\circ} - \angle MNI - \angle I_ANM = 180^{\circ} - \angle I_ANI =180^{\circ}-\angle I_ATI. \]

Let $a,b,c$ be complex numbers such that $a^2,b^2,c^2$ are respective vertices of triangle $ABC$ lying on unit circle centered at $0$ and midpoints of arcs $AB,BC,CA$ not containing other vertices of the triangle than at their ends are $-ab,-bc,-ca$. Then (here $i$ not to be confused with imaginary unit) $$n=bc,\ t=2a^2-bc,\ i=-(ab+bc+ca),\ m=\frac{b^2+c^2}{2},\ i_A=ab-bc+ca.$$If two from the points $I_A,M,I,T$ are equal we are done. If none are equal we have the expression with no division or multiplying by $0$:$$\frac{m-i}{i_A-i}\cdot\frac{t-i_A}{t-m}=\frac{\frac12(b+c)(2a+b+c)}{2a(b+c)}\cdot\frac{a(2a-b-c)}{\frac12(2a-b-c)(2a+b+c)}=\frac12\in\mathbb{R}$$QED

Let $P$ be the midpoint of the minor arc $\widehat{BC}$. Also, $N'$ denote the reflection of $N$ over $P$. Firstly, note that we have $PI\cdot PI_A=-PI^2=-PB^2=-PM\cdot PN=PM\cdot PN'$. So $II_AMN'$ is cyclic. Now to finish, note that the center of $\odot(II_AMN')$ lies on the perpendicular bisector of $II_A$. So the circle $\odot(II_AMN')$ is fixed over the reflection w.r.t. the perpendicular bisector of $II_A$. Now since $PT=PN=PN'$, we get that $\measuredangle N'TN=90^\circ$. Also note that from midpoint theorem, we get that the line perpendicular to $AI$ at $P$ is the perpendicular bisector of $TN'$. So now upon reflecting over the perpendicular bisector of $II_A$, note that $N'\mapsto T$ and we are done.