Solution:

We show that $T$ and $H$ are both on the angle bisector $\ell$ of $\angle BSC$.

We first prove that $H \in \ell$. The altitude $CH$ in triangle $ABC$ is also the altitude in isosceles triangle $PBC$ with $CP = CB$. Therefore, $CH$ is also the angle bisector of $\angle PCB$ and hence also of $\angle SCB$. Analogously, $BH$ is the angle bisector of $\angle SBC$. We conclude that $H$, as the intersection of two angle bisectors of $\triangle BSC$, is also on the third angle bisector, which is $\ell$.

We now prove that $T \in \ell$.

Variant 1. In the isosceles triangles $\triangle BCP$ and $\triangle CBQ$ we see that $\angle BCP = 180^\circ - 2\angle B$ and $\angle CBQ = 180^\circ - 2\angle C$. This yields $\angle PSQ = \angle BSC = 180^\circ - (180^\circ - 2\angle B) - (180^\circ - 2\angle C) = 180^\circ - 2\angle A$. Furthermore, $\angle PTQ = 2\angle PAQ = 2\angle A$ ($T$ being circumcentre of $\triangle APQ$). Now $\angle PTQ + \angle PSQ = 180^\circ$, so $PTQS$ is a cyclic quadrilateral. From $PT = TQ$ we then obtain that $\angle PST = \angle PQT = \angle QPT = \angle QST$, so $T$ is on the angle bisector $\angle PSQ$, which is also $\ell$.

We conclude that $T$, $S$ and $H$ are collinear.

Variant 2. Let $R$ be the second intersection of $BQ$ and $\odot APQ$.

$APRQ$ is a cyclic quadrilateral, so $\angle PRS = \angle A$, $\angle APR = \angle BQC = \angle C$. On the other hand $\angle BPC = \angle B$. Therefore $\angle RPS = 180^\circ - \angle B - \angle C = \angle A$. Hence, the triangle $PRS$ is isosceles with $SP = SR$; then $\ell$ is the perpendicular bisector of the chord $PR$ in the circle that passes through $T$.

Note that if $BQ$ is tangent to $\odot APQ$, then $CP$ is also tangent to $\odot APQ$ and triangle $ABC$ is isosceles, so $T$, $S$ and $H$ lie on the altitude from $A$.

Remark. The fact that $CP$ is also tangent to $\odot APQ$ could be shown with $PTQS$ being a cyclic quadrilateral like in the first variant of the solution. Otherwise we can consider $R$ the second intersection of $CP$ and $\odot APQ$ and prove that triangle $QRS$ is isosceles.

Solution:

In the same way as in the previous solution, we see that $\angle PSQ = 180^\circ - 2\angle A$, so $\angle CSQ = 2\angle A$. From the cyclic quadrilateral $AEHD$ (with $E$ and $D$ feet of the altitudes $CH$ and $BH$) we see that $\angle DHC = \angle DAE = \angle A$. Since $BH$ is the perpendicular bisector of $CQ$, we have $\angle DHQ = \angle A$ as well, so $\angle CHQ = 2\angle A$. From $\angle CHQ = 2\angle A = \angle CSQ$, we see $CHSQ$ is a cyclic quadrilateral. This means $\angle QHS = \angle QCS$.

Since triangles $PTQ$ and $CHQ$ are both isosceles with apex $2\angle A$, we get $\triangle PTQ \sim \triangle CHQ$. We see that one can be obtained from the other by a spiral similarity centered at $Q$, so we also obtain $\triangle QTH \sim \triangle QPC$. This means that $\angle QHT = \angle QCP$. Combining this with $\angle QHS = \angle QCS$, we see that $\angle QHT = \angle QCP = \angle QCS = \angle QHS$. So $\angle QHT = \angle QHS$, which means that $T$, $S$ and $H$ are collinear.

Solution:

Let us draw a parallel $f$ to $BC$ through $A$. Let $B' = BQ \cap f$, $C' = CP \cap f$. Then $AQB' \sim CQB$ and $APC' \sim BPC$, therefore both $QAB'$ and $APC'$ will be isosceles.

Also, $BCS \sim B'C'S$ with respect to the similarity with center $S$, therefore if we take the image of the line $BH$ (which is the perpendicular bisector of the segment $CQ$) through this transformation, it will go through the point $B'$, and be perpendicular to $AQ$ (as the image of $CQ$ is parallel to $CQ$). As $AQB'$ is isosceles, this line is the perpendicular bisector of the segment $AQ$. This means it goes through $T$, the circumcenter of $APQ$. Similarly on the other side the image of $CH$ also goes through $T$. This means that the image of $H$ with respect to the similarity through $S$ will be $T$, so $T$, $S$, $H$ are collinear.