14. Given the function $f(x)=\sin ^{4} x$.
(1) Let $g(x)=f(x)+f\left(\frac{\pi}{2}-x\right)$, find the maximum and minimum values of $g(x)$ in the interval $\left[\frac{\pi}{6}, \frac{3 \pi}{8}\right]$;
(2) Find the value of $\sum_{k=1}^{89} f\left(\frac{k \pi}{180}\right)$.
$$
\begin{array}{l}
(1) \text{ From the given,} \\
g(x)=\sin ^{4} x+\cos ^{4} x \\
=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cdot \cos ^{2} x \\
=1-\frac{1}{2} \sin ^{2} 2 x. \\
\text{ Given } x \in\left[\frac{\pi}{6}, \frac{3 \pi}{8}\right] \Rightarrow 2 x \in\left[\frac{\pi}{3}, \frac{3 \pi}{4}\right] \\
\Rightarrow \sin 2 x \in\left[\frac{\sqrt{2}}{2}, 1\right] \\
\Rightarrow g(x) \in\left[\frac{1}{2}, \frac{3}{4}\right]. \\
\end{array}
$$

Therefore, the maximum value of $g(x)$ in the interval $\left[\frac{\pi}{6}, \frac{3 \pi}{8}\right]$ is $g\left(\frac{3 \pi}{8}\right)=\frac{3}{4}$, and the minimum value is $g\left(\frac{\pi}{4}\right)=\frac{1}{2}$.
$$
\begin{array}{l}
(2) \sum_{k=1}^{88} f\left(\frac{k \pi}{180}\right) \\
=\sum_{k=1}^{44} g\left(\frac{k \pi}{180}\right)+\sin ^{4} \frac{45 \pi}{180} \\
=44-\frac{1}{2} \sum_{k=1}^{88} \sin ^{2} \frac{k \pi}{180}+\frac{1}{4} \\
=44-\frac{1}{2} \times 22+\frac{1}{4}=\frac{133}{4}.
\end{array}
$$