I liked G4 and I couldn't resist being able to advertise local contests in my solution. Let $J$ be the $A$-excenter and let $I^*$ be the reflection of $I$ about $BC$. By Mexico National 2015 P5 we get that $D, I^*, J$ are collinear. Moreover, $J$ lies on $(BIC)$, so upon reflecting about $BC$ we find that the reflection of $J$ about $BC$ is the intersection of $DI$ and the reflection of $(BIC)$ about $BC$. This reflected circle is clearly $(BI'C)$, so $T$ is the reflection of $J$ about $BC$. Moreover notice that if $D_1$ is the projection of $I'$ onto $BC$ then $BD = CD_1$, which implies that $J, I', X, T$ are all collinear and $TX \parallel AD$. Now let $Y = AP \cap BC, Z = AD \cap PI$ and notice that $PYDZ$ is cyclic as $\angle ZPY = \angle ZDY = 90^{\circ}$. Moreover by Radical Axis with $(API), (ABC), (BIC)$ we see that $Y$ lies on the radical axis of $(API)$ and $(BIC)$ which is the tangent to $(API)$ at $I$ as the circumcenters of these triangles both lie on $AI$. We now easily get $AI^2 = AP \cdot AY = AZ \cdot AD$ by Power of a Point from $A$ to the circumcircles of $(IPY)$ and $(DZPY)$. Thus $AI$ is tangent to the circumcircle of $IZD$ and therefore $$\angle AIP = \angle AIZ = \angle ADI = \angle ITX = \angle IAX$$ Which completes the problem.

Here is another solution to G4. As with juckter we get easily that $T$ is the reflection of $I_A$ over $BC$. Let $N$ be the midpoint of arc $BC$ of $(ABC)$ and let $D_0$ be the foot from $I$ to $BC$. It is well known (e.g. invert about the incircle) that $P, D_0, N$ are collinear, and now we have that $PEFI \sim PBCN$ by Spiral Sim so if $M$ is the midpoint of $BC$ we have \[ \angle PIA = \angle PNM = \angle D_0NM = \angle D_1NM \]where $D_1$ is the extouch point. Let $PI$ meet $TX$ at $Q$, then \[ \angle ND_1Q = \angle MND_1 = \angle PIA = \angle QIN, \]so $(INQD_1)$ is cyclic. In particular, \[ \angle IQT = \angle IND_1 = \angle AIT = \angle AXT, \]so $IQ\parallel AX$. Here we use the fact that $IT\parallel NX$ by homothety centered at $I_A$ with ratio $2$. $\blacksquare$