Given that the maximum height reached by the first ball is 8 times that of the second ball:

$ \left(H_{\max}\right)_1 = 8 \times \left(H_{\max}\right)_2 $

We can relate this to the initial velocities and angles using the formula for maximum height:

$ \frac{u^2 \sin^2 \theta_1}{2g} = 8 \times \frac{u^2 \sin^2 \theta_2}{2g} $

Simplifying, we find:

$ \sin \theta_1 = 2 \sqrt{2} \sin \theta_2 $

Now, the ratio of the total flying times $T_1$ and $T_2$ is given by:

$ \frac{T_1}{T_2} = \frac{2u \sin \theta_1 / g}{2u \sin \theta_2 / g} = \frac{\sin \theta_1}{\sin \theta_2} = 2 \sqrt{2} $

Thus, the ratio of $T_1$ to $T_2$ is $2 \sqrt{2} : 1$.