If $$a = \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^4}} - \sqrt{2}}{x^4}$$ and $$b = \lim_{x \to 0} \frac{\sin^2 x}{\sqrt{2} - \sqrt{1 + \cos x}}$$, then the value of $$ab^3$$ is :

Finding $$a = \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^4}} - \sqrt{2}}{x^4}$$:

Let $$u = x^4 \to 0$$. Then:

$$a = \lim_{u \to 0} \frac{\sqrt{1 + \sqrt{1+u}} - \sqrt{2}}{u}$$

Rationalizing: multiply by $$\frac{\sqrt{1+\sqrt{1+u}} + \sqrt{2}}{\sqrt{1+\sqrt{1+u}} + \sqrt{2}}$$:

$$a = \lim_{u \to 0} \frac{1 + \sqrt{1+u} - 2}{u(\sqrt{1+\sqrt{1+u}} + \sqrt{2})} = \lim_{u \to 0} \frac{\sqrt{1+u} - 1}{u} \cdot \frac{1}{\sqrt{1+\sqrt{1+u}} + \sqrt{2}}$$

$$\lim_{u \to 0} \frac{\sqrt{1+u}-1}{u} = \frac{1}{2}$$ and $$\sqrt{1+\sqrt{1+0}} + \sqrt{2} = 2\sqrt{2}$$.

$$a = \frac{1}{2} \cdot \frac{1}{2\sqrt{2}} = \frac{1}{4\sqrt{2}}$$

Finding $$b = \lim_{x \to 0} \frac{\sin^2 x}{\sqrt{2} - \sqrt{1 + \cos x}}$$:

Rationalizing: multiply by $$\frac{\sqrt{2} + \sqrt{1 + \cos x}}{\sqrt{2} + \sqrt{1 + \cos x}}$$:

$$b = \lim_{x \to 0} \frac{\sin^2 x(\sqrt{2} + \sqrt{1 + \cos x})}{2 - (1 + \cos x)} = \lim_{x \to 0} \frac{\sin^2 x(\sqrt{2} + \sqrt{1+\cos x})}{1 - \cos x}$$

Using $$\frac{\sin^2 x}{1 - \cos x} = \frac{1 - \cos^2 x}{1 - \cos x} = 1 + \cos x$$:

$$b = \lim_{x \to 0} (1 + \cos x)(\sqrt{2} + \sqrt{1 + \cos x}) = 2(\sqrt{2} + \sqrt{2}) = 2 \cdot 2\sqrt{2} = 4\sqrt{2}$$

$$ab^3 = \frac{1}{4\sqrt{2}} \times (4\sqrt{2})^3 = \frac{1}{4\sqrt{2}} \times 64 \times 2\sqrt{2} = \frac{128\sqrt{2}}{4\sqrt{2}} = 32$$

The answer is $$\boxed{32}$$, which corresponds to Option (2).