Solution:

a)
Since the circles $k_{1}$ and $k_{2}$ are tangent at the point $T$, we have
$$\text{B X T = X T 2 = T S 2 = T A S.}$$
Then it follows easily that $S$ is the midpoint of the arc $\widehat{A B}$, i.e. $S A = S B$. Hence $\text{T C I = T A S}$ (the quadrilateral $A T C S$ is cyclic), $\text{T A S = B X T}$ and $\text{B X T = T Y X}$. Therefore $\text{T C I = T Y I}$, which shows that the quadrilateral $C T Y I$ is cyclic.

b)
Since $\text{A X S = T A S}$ it follows easily that $\triangle A X S \sim \triangle T A S$, whence $S A^{2} = S T \cdot S X$. We have from a) that $\text{C I T = C Y T = T X Y}$ and therefore $\triangle S X I \sim \triangle S I T$, whence $S I^{2} = S T \cdot S X$. Hence $S A = S I$. On the other hand, it follows from
$$\text{B C I = 180 - B C S = 180 - ( + + 2 ) = 90 - 2}$$
that $C I$ is the external bisector of $\text{A C B}$.

In the isosceles $\triangle B S I$ we have $\text{B S I = B S C =}$ and we find $\text{B I S = 90 - 2}$. Now from $\triangle B C I$ we have $\text{C B I = 90 - 2}$, which means that $B I$ is the external bisector of $\text{A B C}$. Therefore $I$ is the center of the excircle of $\triangle A B C$ tangent to the side $B C$.