Statement I: The equation $$(\sin^{-1}x)^3 + (\cos^{-1}x)^3 - a\pi^3 = 0$$ has a solution for all $$a \geq \frac{1}{32}$$. 

Statement II: For any x $$\in$$ R, $$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$$ and $$0 \leq \left(\sin^{-1}x - \frac{\pi}{4}\right)^2 \leq \frac{9\pi^2}{16}$$.

Let $$f(x) = (\sin^{-1}x)^3 + (\cos^{-1}x)^3$$. Using the identity $$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$$ for $$x \in [-1, 1]$$, let $$\sin^{-1}x = t$$, where $$t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$$.

$$f(x) = t^3 + \left(\frac{\pi}{2} - t\right)^3$$

$$A^3 + B^3 = (A+B)(A^2 - AB + B^2)$$: 

$$f(x) = \left(t + \frac{\pi}{2} - t\right) \left[t^2 - t\left(\frac{\pi}{2} - t\right) + \left(\frac{\pi}{2} - t\right)^2\right]$$

$$f(x) = \frac{\pi}{2} \left[t^2 - \frac{\pi t}{2} + t^2 + \frac{\pi^2}{4} - \pi t + t^2\right]$$

$$f(x) = \frac{\pi}{2} \left[3t^2 - \frac{3\pi t}{2} + \frac{\pi^2}{4}\right]$$

$$f(x) = \frac{3\pi}{2} \left[t^2 - \frac{\pi t}{2} + \frac{\pi^2}{12}\right]$$

$$f(x) = \frac{3\pi}{2} \left[\left(t - \frac{\pi}{4}\right)^2 + \frac{\pi^2}{48}\right]$$

For $$x \in [-1, 1]$$, the range of $$t = \sin^{-1}x$$ is $$[-\frac{\pi}{2}, \frac{\pi}{2}]$$.

The range of $$\left(t - \frac{\pi}{4}\right)$$ is $$[-\frac{3\pi}{4}, \frac{\pi}{4}]$$.

Squaring this interval: $$0 \le \left(\sin^{-1}x - \frac{\pi}{4}\right)^2 \le \left(-\frac{3\pi}{4}\right)^2 = \frac{9\pi^2}{16}$$

Thus, Statement II is true.

Using the range of $$\left(t - \frac{\pi}{4}\right)^2$$ in the expression for $$f(x)$$:

Minimum value: $$f(x)_{min} = \frac{3\pi}{2} \left[0 + \frac{\pi^2}{48}\right] = \frac{\pi^3}{32}$$

Maximum value: $$f(x)_{max} = \frac{3\pi}{2} \left[\frac{9\pi^2}{16} + \frac{\pi^2}{48}\right] = \frac{3\pi}{2} \left[\frac{28\pi^2}{48}\right] = \frac{7\pi^3}{8}$$

The given equation $$(\sin^{-1}x)^3 + (\cos^{-1}x)^3 = a\pi^3$$ has a solution if $$a\pi^3 \in [\frac{\pi^3}{32}, \frac{7\pi^3}{8}]$$.

This implies $$\frac{1}{32} \le a \le \frac{7}{8}$$

In the context of the given statements, the condition $$a \ge \frac{1}{32}$$ identifies the lower bound necessary for a solution to exist. Hence, Statement I is false.