Solution:

Answer: $\frac{3}{5}$

Since the problem only deals with ratios, we can assume that the radius of $O$ is $1$. As we have proven in Problem 5, points $S_1$ and $S_2$ are midpoints of arc $AB$. Since $AB$ is a diameter, $S_1 S_2$ is also a diameter, and thus $S_1 S_2 = 2$.

Let $O_1$, $O_2$, and $P$ denote the centers of circles $\omega_1$, $\omega_2$, and $O$. Since $\omega_1$ is tangent to $O$, we have $PO_1 + O_1 X = 1$. But $O_1 X \perp AB$. So $\triangle PO_1 X$ is a right triangle, and $O_1 X^2 + XP^2 = O_1 P^2$. Thus, $O_1 X^2 + 1/4 = (1 - O_1 X)^2$, which means $O_1 X = \frac{3}{8}$ and $O_1 P = \frac{5}{8}$.

Since $T_1 T_2 \parallel O_1 O_2$, we have $T_1 T_2 = O_1 O_2 \cdot \frac{PT_1}{PO_1} = 2 O_1 X \cdot \frac{PT_1}{PO_1} = 2 \left(\frac{3}{8}\right) \frac{1}{5/8} = \frac{6}{5}$. Thus $\frac{T_1 T_2}{S_1 S_2} = \frac{6/5}{2} = \frac{3}{5}$.