(1) The smallest positive period of the function $f(x)=\cos ^{4} x+\sin ^{2} x(x \in \mathbf{R})$ is ( ).
(A) $\frac{\pi}{4}$
(B) $\frac{\pi}{2}$
(C) $\pi$
(D) $2 \pi$
1 Method one:
$$
\begin{aligned}
f\left(x+\frac{\pi}{2}\right) & =\sin ^{4} x+\cos ^{2} x \\
& =\sin ^{4} x+\cos ^{4} x+\cos ^{2} x \sin ^{2} x \\
& =\cos ^{4} x+\sin ^{2} x \\
& =f(x),
\end{aligned}
$$

Since $f(0)=1, f\left(\frac{\pi}{4}\right)=\frac{1}{4}+\frac{1}{2} \neq f(0)$, the correct choice is B.
Method two:
$$
\begin{aligned}
f(x) & =\cos ^{4} x+1-\cos ^{2} x \\
& =\cos ^{2} x\left(\cos ^{2} x-1\right)+1 \\
& =1-\cos ^{2} x \sin ^{2} x \\
& =1-\frac{1}{4} \sin ^{2} 2 x \\
& =\frac{1}{8} \cos 4 x+\frac{7}{8},
\end{aligned}
$$

It is known that the smallest positive period of $f(x)$ is $\frac{2 \pi}{4}=\frac{\pi}{2}$. Therefore, the correct choice is B.