1. Compute $a$

We need

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

Let

$$y=\sqrt{1+x^4}.$$

Then as $x\to 0$, $y\to 1$, and

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

Rationalize the numerator:

$$\sqrt{1+y}-\sqrt{2}=\frac{(1+y)-2}{\sqrt{1+y}+\sqrt{2}}=\frac{y-1}{\sqrt{1+y}+\sqrt{2}}.$$

So,

$$a=\lim_{x\to 0}\frac{\sqrt{1+x^4}-1}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)}.$$

Now rationalize $\sqrt{1+x^4}-1$:

$$\sqrt{1+x^4}-1=\frac{x^4}{\sqrt{1+x^4}+1}.$$

Hence,

$$a=\lim_{x\to 0}\frac{1}{(\sqrt{1+x^4}+1)(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})}.$$

Putting $x=0$,

$$a=\frac{1}{(1+1)(\sqrt{2}+\sqrt{2})}=\frac{1}{2\cdot 2\sqrt{2}}=\frac{1}{4\sqrt{2}}.$$

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2. Compute $b$

We need

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

Rationalize the denominator:

$$\sqrt{2}-\sqrt{1+\cos x} =\frac{2-(1+\cos x)}{\sqrt{2}+\sqrt{1+\cos x}} =\frac{1-\cos x}{\sqrt{2}+\sqrt{1+\cos x}}.$$

Therefore,

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

Use

$$\sin^2 x=(1-\cos x)(1+\cos x).$$

So,

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

Now substitute $x=0$:

$$b=(1+1)(\sqrt{2}+\sqrt{2})=2\cdot 2\sqrt{2}=4\sqrt{2}.$$

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3. Find $ab^3$

We have

$$a=\frac{1}{4\sqrt{2}},\qquad b=4\sqrt{2}.$$

Then

$$b^3=(4\sqrt{2})^3=64\cdot 2\sqrt{2}=128\sqrt{2}.$$

Thus,

$$ab^3=\frac{1}{4\sqrt{2}}\cdot 128\sqrt{2}=\frac{128}{4}=32.$$

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4. Option check

The value is

$$\boxed{32}.$$

So the correct option is C.