1. Recall the definition of periodicity

A function $f(x)$ is periodic if there exists some $T>0$ such that $f(x+T)=f(x)$ for all $x$ in its domain.

We check each option.

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2. Option A: $\left|\sin 3x\right|+\sin^2 x$

• $\left|\sin 3x\right|$ is periodic. Since $\sin 3x$ has period $\dfrac{2\pi}{3}$, its absolute value has period $\frac{\pi}{3}$ because |\sin(\theta+\pi)|=|-sin\theta|=|\sin\theta|.

• $\sin^2 x$ has period $\pi$

Now we check if these periods are commensurable: $\pi = 3\left(\frac{\pi}{3}\right)$ So a common period exists, namely $\pi$.

Hence option A is periodic.

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3. Option B: $\cos\sqrt{x}+\cos^2 x$

• $\cos^2 x$ is periodic with period $\pi$.
• But $\cos\sqrt{x}$ is not periodic.

Let us justify this.

If $\cos\sqrt{x}$ were periodic with period $T>0$, then for all $x\ge 0$, $\cos\sqrt{x+T}=\cos\sqrt{x}.$ This would require the argument difference to behave like a constant multiple of $2\pi$, but $\sqrt{x+T}-\sqrt{x}=\frac{T}{\sqrt{x+T}+\sqrt{x}}$ which depends on $x$ and tends to $0$ as $x\to\infty$. So no fixed $T>0$ can make $\cos\sqrt{x+T}=\cos\sqrt{x}$ for all $x$.

Therefore $\cos\sqrt{x}$ is not periodic, and hence the sum $\cos\sqrt{x}+\cos^2 x$ is not periodic.

So option B is not periodic.

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4. Option C: $\cos 4x+\tan^2 x$

• $\cos 4x$ has period $\frac{\pi}{2}$
• $\tan x$ has period $\pi$, so $\tan^2 x$ has period $\pi$ (in fact also not smaller than $\pi$ as a fundamental period).

A common period exists: $\pi = 2\left(\frac{\pi}{2}\right).$ So option C is periodic with period $\pi$.

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5. Option D: $\cos 2x+\sin x$

• $\cos 2x$ has period $\pi$.
• $\sin x$ has period $2\pi$.

A common period exists, namely $2\pi$. Hence option D is periodic.

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6. Conclusion

The only function which is not periodic is $\boxed{\cos\sqrt{x}+\cos^2 x}$ so the correct option is $\boxed{B}$

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7. Comparison with stored correct answer

Stored correct answer: $B$

Our derived answer is also $B$, so they agree.