Let's analyze the problem step-by-step.

Energy conservation:

When the sphere rolls without slipping, its gravitational potential energy converts into both translational and rotational kinetic energy. The energy conservation equation is given by:

$$mgL\sin\theta = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

where:

$L\sin\theta$ is the vertical height,

$I$ is the moment of inertia, and

$\omega$ is the angular speed.

Moment of Inertia and Rolling Condition:

For a solid sphere, the moment of inertia about its center is:

$$I = \frac{2}{5}mr^2$$

Since the sphere rolls without slipping, the linear speed $v$ and the angular speed $\omega$ are related by:

$$v = r\omega \quad \Rightarrow \quad \omega = \frac{v}{r}$$

Total Kinetic Energy:

Substituting the rolling condition into the rotational kinetic energy gives:

$ \begin{aligned} K &= \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right) \left(\frac{v}{r}\right)^2 \\ &= \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right) \frac{v^2}{r^2} \\ &= \frac{1}{2}mv^2 + \frac{1}{5}mv^2 \\ &= \frac{7}{10}mv^2 \end{aligned} $

Finding $v^2$ in Terms of $L$ and $\sin\theta$:

Equate the initial potential energy to the final kinetic energy:

$$mgL\sin\theta = \frac{7}{10}mv^2$$

Solving for $v^2$:

$$v^2 = \frac{10}{7}gL\sin\theta$$

Apply to the Two Cases:

For $\theta = 30^\circ$, since $\sin(30^\circ) = \frac{1}{2}$:

$$v_1^2 = \frac{10}{7}gL\left(\frac{1}{2}\right) = \frac{5}{7}gL$$

For $\theta = 45^\circ$, since $\sin(45^\circ) = \frac{1}{\sqrt{2}}$:

$$v_2^2 = \frac{10}{7}gL\left(\frac{1}{\sqrt{2}}\right)$$

Calculate the Ratio $\frac{v_1^2}{v_2^2}$:

$ \begin{aligned} \frac{v_1^2}{v_2^2} &= \frac{\left(\frac{5}{7}gL\right)}{\left(\frac{10}{7}gL\frac{1}{\sqrt{2}}\right)} \\ &= \frac{5}{10}\sqrt{2} \\ &= \frac{\sqrt{2}}{2} \end{aligned} $

Expressing this ratio as $v_1^2 : v_2^2$, we have:

$$v_1^2 : v_2^2 = 1 : \sqrt{2}$$

Conclusion:

According to the options given, the correct answer is:

Option C: $1:\sqrt{2}$.