(2) Given that $\theta$ is an angle in the third quadrant, and $\sin \frac{\theta}{4}<\cos \frac{\theta}{4}$, then the range of $\frac{\theta}{4}$ is
(A)
$$
\begin{array}{l}
\left(2 k \pi+\frac{\pi}{4}, 2 k \pi+\frac{3 \pi}{8}\right) \cup\left(2 k \pi+\frac{3}{4} \pi, 2 k \pi+\frac{7}{8} \pi\right), \\
k \in \mathbf{Z}
\end{array}
$$
(B) $\left(2 k \pi+\frac{5}{4} \pi, 2 k \pi+\frac{11}{8} \pi\right) \cup\left(2 k \pi+\frac{7}{4} \pi, 2 k \pi+\frac{15}{8} \pi\right)$, $k \in \mathbf{Z}$
(C) $\left(k \pi+\frac{\pi}{4}, k \pi+\frac{3 \pi}{8}\right) \cup\left(k \pi+\frac{3 \pi}{4}, k \pi+\frac{7}{8} \pi\right), k \in \mathbf{Z}$
(D) $\left(4 k \pi+\frac{5}{4} \pi, 4 k \pi+\frac{11}{8} \pi\right) \cup\left(4 k \pi+\frac{7}{4} \pi, 4 k \pi+\frac{15}{8} \pi\right)$, $k \in \mathbf{Z}$
(2) B Hint: Since $2 k \pi+\pi<\theta<2 k \pi+\frac{3 \pi}{2}$, then $\frac{k \pi}{2}+\frac{\pi}{4}<\frac{\theta}{4}<\frac{k \pi}{2}+\frac{3 \pi}{8}, k \in \mathbf{Z}$.

Therefore, the region where $\frac{\theta}{4}$ is located is the shaded part shown in the figure.

Also, because $\sin \frac{\theta}{4}<\cos \frac{\theta}{4}$, the range of $\frac{\theta}{4}$ is
$$
\left(2 k \pi+\frac{5}{4} \pi, 2 k \pi+\frac{11}{8} \pi\right) \cup\left(2 k \pi+\frac{7}{4} \pi, 2 k \pi+\frac{15}{8} \pi\right), k \in \mathbf{Z}
$$