1. Interpret the function

The given function is $f(x)=\sin 4x+\cos 4x.$

To find where $f$ is increasing, we compute its derivative and check where $f'(x)>0.$

2. Differentiate

$f'(x)=4\cos 4x-4\sin 4x=4(\cos 4x-\sin 4x).$

So $f$ is increasing when $\cos 4x-\sin 4x>0.$

3. Simplify the inequality

Use the identity $\cos \theta-\sin \theta=\sqrt{2}\cos\left(\theta+\frac{\pi}{4}\right).$

Thus, $f'(x)=4\sqrt{2}\cos\left(4x+\frac{\pi}{4}\right).$

Hence $f$ is increasing when $\cos\left(4x+\frac{\pi}{4}\right)>0.$

Now, $\cos y>0$ for $y\in \left(-\frac{\pi}{2}+2n\pi,\; \frac{\pi}{2}+2n\pi\right), \quad n\in \mathbb{Z}.$

So, $4x+\frac{\pi}{4}\in \left(-\frac{\pi}{2}+2n\pi,\; \frac{\pi}{2}+2n\pi\right).$

Subtract $\frac{\pi}{4}$: $4x\in \left(-\frac{3\pi}{4}+2n\pi,\; \frac{\pi}{4}+2n\pi\right).$

Divide by $4$: $x\in \left(-\frac{3\pi}{16}+\frac{n\pi}{2},\; \frac{\pi}{16}+\frac{n\pi}{2}\right).$

4. Check the given options

Now test each interval.

Option A: $\left(0,\frac{\pi}{4}\right)$

This entire interval is not contained in any increasing interval above. For example, at $x=\frac{\pi}{8}$, $f'\left(\frac{\pi}{8}\right)=4(\cos \tfrac{\pi}{2}-\sin \tfrac{\pi}{2})=4(0-1)=-4<0.$ So A is not correct.

Option B: $\left(\frac{\pi}{4},\frac{\pi}{2}\right)$

Take $n=1$ in the general interval: $x\in \left(-\frac{3\pi}{16}+\frac\pi2,\; \frac{\pi}{16}+\frac\pi2\right)=\left(\frac{5\pi}{16},\; \frac{9\pi}{16}\right).$ This is the increasing interval near option B. But option B is $\left(\frac{\pi}{4},\frac{\pi}{2}\right)=\left(\frac{4\pi}{16},\frac{8\pi}{16}\right).$ This is not fully contained in $\left(\frac{5\pi}{16},\frac{9\pi}{16}\right)$, since points between $\frac{\pi}{4}$ and $\frac{5\pi}{16}$ are not increasing. For example, at $x=\frac{\pi}{4}$ (or just to the right of it), $f'\left(\frac{\pi}{4}\right)=4(\cos \pi-\sin \pi)=4(-1-0)=-4<0.$ So B is not correct as a whole interval.

Option C: $\left(\frac{\pi}{2},\frac{5\pi}{8}\right)$

This is $\left(\frac{8\pi}{16},\frac{10\pi}{16}\right),$ which is not contained in $\left(\frac{5\pi}{16},\frac{9\pi}{16}\right).$ Also, for $x=\frac{19\pi}{32}\in \left(\frac\pi2,\frac{5\pi}{8}\right)$, $4x=\frac{19\pi}{8}$, so derivative changes sign within the interval. Hence C is not correct.

Option D: $\left(\frac{5\pi}{8},\frac{3\pi}{4}\right)$

Take $n=2$: $x\in \left(-\frac{3\pi}{16}+\pi,\; \frac{\pi}{16}+\pi\right),$ which is beyond this region, so not relevant. Also around $x=\frac{11\pi}{16}$, $f'(x)=4(\cos \tfrac{11\pi}{4}-\sin \tfrac{11\pi}{4})<0,$ so D is not correct.

5. Conclusion

The exact intervals where $f$ is increasing are $\boxed{x\in \left(-\frac{3\pi}{16}+\frac{n\pi}{2},\; \frac{\pi}{16}+\frac{n\pi}{2}\right),\; n\in\mathbb Z.}$

Among the given options, none matches completely.

So the stored answer B is not correct for the function as written.

Note: If the intended function were $f(x)=\sin^4 x+\cos^4 x$, then the answer would be different. But for $f(x)=\sin 4x+\cos 4x$, none of the options is fully correct.