A possible value of $$\tan\left(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\right)$$ is:

We need to find $$\tan\left(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\right)$$.

Let $$\theta = \sin^{-1}\frac{\sqrt{63}}{8}$$, so $$\sin\theta = \frac{\sqrt{63}}{8}$$ and $$\cos\theta = \sqrt{1 - \frac{63}{64}} = \frac{1}{8}$$.

Using the half-angle formula, $$\cos\frac{\theta}{2} = \sqrt{\frac{1 + \cos\theta}{2}} = \sqrt{\frac{1 + \frac{1}{8}}{2}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$$.

Similarly, $$\sin\frac{\theta}{2} = \sqrt{\frac{1 - \cos\theta}{2}} = \sqrt{\frac{1 - \frac{1}{8}}{2}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$$.

Now we apply the half-angle formula again to find $$\tan\frac{\theta}{4}$$. Using $$\tan\frac{\alpha}{2} = \frac{1 - \cos\alpha}{\sin\alpha}$$ with $$\alpha = \frac{\theta}{2}$$, we get $$\tan\frac{\theta}{4} = \frac{1 - \cos\frac{\theta}{2}}{\sin\frac{\theta}{2}} = \frac{1 - \frac{3}{4}}{\frac{\sqrt{7}}{4}} = \frac{\frac{1}{4}}{\frac{\sqrt{7}}{4}} = \frac{1}{\sqrt{7}}$$.

Therefore, a possible value of $$\tan\left(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\right)$$ is $$\frac{1}{\sqrt{7}}$$.