7. Let $f(x)=\sin ^{4} x-\sin x \cdot \cos x+\cos ^{4} x$. Then the range of $f(x)$ is $\qquad$ .
$$
\begin{array}{l}
\text { II.7.[0, } \left.\frac{9}{8}\right] . \\
f(x)=\sin ^{4} x-\sin x \cdot \cos x+\cos ^{4} x \\
=1-\frac{1}{2} \sin 2 x-\frac{1}{2} \sin ^{2} 2 x
\end{array}
$$

Let $t=\sin 2 x$, then
$$
\begin{array}{l}
f(x)=g(t) \\
=1-\frac{1}{2} t-\frac{1}{2} t^{2}=\frac{9}{8}-\frac{1}{2}\left(t+\frac{1}{2}\right)^{2} .
\end{array}
$$

Therefore, $\min _{-1 \leqslant t \leqslant 1} g(t)=g(1)=\frac{9}{8}-\frac{1}{2} \times \frac{9}{4}=0$,
$$
\max _{-1<t<1} g(t)=g\left(-\frac{1}{2}\right)=\frac{9}{8}-\frac{1}{2} \times 0=\frac{9}{8} \text {. }
$$

Thus, $0 \leqslant f(x) \leqslant \frac{9}{8}$.