1. Find the intersection points of $y=\sin x$ and $y=\cos x$

We solve $\sin x = \cos x$ which gives $\tan x = 1$ So, $x=\frac{\pi}{4}+n\pi, \quad n\in \mathbb{Z}.$

Thus, two consecutive intersection points are at $x=\frac{\pi}{4} \quad \text{and} \quad x=\frac{5\pi}{4}.$

2. Determine which curve is above the other in this interval

Take a test point, say $x=\pi$: $\sin \pi = 0, \qquad \cos \pi = -1.$ Hence, $\sin x > \cos x$ on $\left[\frac{\pi}{4},\frac{5\pi}{4}\right]$.

Therefore, the enclosed area is $A=\int_{\pi/4}^{5\pi/4}(\sin x-\cos x)\,dx.$

3. Evaluate the integral

$A=\int_{\pi/4}^{5\pi/4}(\sin x-\cos x)\,dx$ $=\left[-\cos x-\sin x\right]_{\pi/4}^{5\pi/4}.$

Now, $\cos\frac{\pi}{4}=\sin\frac{\pi}{4}=\frac{\sqrt2}{2},$ so $-\cos\frac{\pi}{4}-\sin\frac{\pi}{4}=-\sqrt2.$

Also, $\cos\frac{5\pi}{4}=\sin\frac{5\pi}{4}=-\frac{\sqrt2}{2},$ so $-\cos\frac{5\pi}{4}-\sin\frac{5\pi}{4}=\sqrt2.$

Hence, $A=\sqrt2-(-\sqrt2)=2\sqrt2.$

4. Compute $A^4$

$A^4=(2\sqrt2)^4.$ Now, $2\sqrt2=2^{3/2}$ so $(2\sqrt2)^4=(2^{3/2})^4=2^6=64.$

Therefore, $\boxed{A^4=64}.$