(1) Draw auxiliary lines as shown in the figure. Since arc $ST =$ arc $XT$, then
$$\angle BXT = \frac{\text{arc}XT^{\circ}}{2} = \frac{\text{arc}ST^{\circ}}{2} = \angle TAS.$$
Hence, $\triangle SAT \sim \triangle SXA$. Therefore $\angle XAS = \angle ATS$. That is, arc $BS =$ arc $AS$.
So $S$ is the midpoint of arc $AB$. Since
$$\angle TCI = \angle TAS = \angle BXT = \angle TYX.$$
Therefore $C, T, I, Y$ are concyclic.

(2) Since $\triangle SAT \sim \triangle SXA$, we have $SA^2 = ST \cdot SX$. Also, since $C, T, I, Y$ are concyclic, we have
$$\angle CIT = \angle CYT = \angle TXY$$
Thus, $\triangle SXI \sim \triangle SIT$.
Therefore, $SI^2 = ST \cdot SX$. So $SA = SI$.
Let $\angle BAC = \alpha, \angle ABC = \beta, \angle ACB = \gamma$, then
$$\angle ACS = \frac{\text{arcAS}^{\circ}}{2} = \frac{\text{arcASB}^{\circ}}{4} = \frac{1}{4}(2\pi - 2\gamma) = \frac{\pi}{2} - \frac{\gamma}{2}$$
$$\text{Hence } \angle BCI = \pi - \angle BCS = \pi - \left(\gamma + \frac{\pi}{2} - \frac{\gamma}{2}\right) = \frac{\pi}{2} - \frac{\gamma}{2}$$

Therefore, $CI$ is the external angle bisector of $\angle ACB$.
Also, since $SB = SA = SI$, $\angle BSI = \angle BSC = \alpha$, we have
$$\angle BIS = \frac{\pi - \angle BSI}{2} = \frac{\pi}{2} - \frac{\alpha}{2}$$
In $\triangle BCI$, we obtain $\angle CBI = \frac{\pi}{2} - \frac{\beta}{2}$, that is, $BI$ is the external angle bisector of $\angle ABC$, so, point $I$ is the center of the excircle opposite $\angle A$ of $\triangle ABC$.