7. In $\triangle A B C$, the three interior angles $A, B, C$ satisfy $A+C=2 B, \frac{1}{\cos A}+\frac{1}{\cos C}=$ $\frac{\sqrt{2}}{\cos B}, \cos \frac{A-C}{2}=$ $\qquad$ .
$\begin{array}{l}\text { 7. }-\frac{\sqrt{2}}{2} \text {. } \\ \left.\begin{array}{l}A+B+C=180^{\circ} \\ A+C=2 B\end{array}\right\} \Rightarrow B=60^{\circ} \cdot \frac{1}{\cos A}+\frac{1}{\cos C}=\frac{\sqrt{2}}{\cos B} \Rightarrow \sqrt{2} \cos A \cos C=(\cos A+ \\ \cos C) \cos B \Rightarrow \frac{\sqrt{2}}{2}(\cos (A+C)+\cos (A-C))=2 \cos \frac{A+C}{2} \cos \frac{A-C}{2} \cos 60^{\circ} \Rightarrow \\ \frac{\sqrt{2}}{2}\left[-\cos B+2 \cos ^{2} \frac{A-C}{2}-1\right]=\frac{1}{2} \cos \frac{A-C}{2} \Rightarrow-\frac{\sqrt{2}}{4}+2 \sqrt{2} \cos ^{2} \frac{A-C}{2}-\frac{1}{2} \cos \frac{A-C}{2}- \\ \frac{\sqrt{2}}{2}=0 \Rightarrow \sqrt{2} \cos ^{2} \frac{A-C}{2}-\frac{1}{2} \cos \frac{A-C}{2}-\frac{3 \sqrt{2}}{4}=0 \Rightarrow \cos ^{2} \frac{A-C}{2}-\frac{1}{2 \sqrt{2}} \cos \frac{A-C}{2}-\frac{3}{4}=0 \Rightarrow \\ \cos \frac{A-C}{2}=\frac{\frac{1}{2 \sqrt{2}} \pm \sqrt{\frac{1}{8}+3}}{2}=\frac{\frac{1}{2 \sqrt{2}} \pm \frac{5}{2 \sqrt{2}}}{2}=\frac{1}{4 \sqrt{2}}(1 \pm 5), \cos \frac{A-C}{2}=-\frac{4}{4 \sqrt{2}}=\frac{\sqrt{2}}{2} \text { or } \\ \cos \frac{A-C}{2}=\frac{3 \sqrt{2}}{4}>1 \text {, i.e., } \cos \frac{A-C}{2}=-\frac{\sqrt{2}}{2} \text {. } \\\end{array}$