7. Given in $\triangle A B C$, $a=2 b, \cos B=\frac{2 \sqrt{2}}{3}$, then $\sin \frac{A-B}{2}+\sin \frac{C}{2}=$ $\qquad$
Let $\sin \frac{A}{2}=m, \cos \frac{A}{2}=n, \sin \frac{B}{2}=x, \cos \frac{B}{2}=y(m, n, x, y>0)$, then $2 y^{2}-1=\cos B=\frac{2 \sqrt{2}}{3} \Rightarrow y^{2}=\frac{2 \sqrt{2}+3}{6} \Rightarrow y=\frac{\sqrt{2}+1}{\sqrt{6}}$, similarly we get $x=\frac{\sqrt{2}-1}{\sqrt{6}} \Rightarrow y-x=\frac{2}{\sqrt{6}}=\frac{\sqrt{6}}{3}$. Also, $a=2 b \Rightarrow \sin A=2 \sin B \Rightarrow m n=2 x y=2 \cdot \frac{1}{6}=\frac{1}{3}$, thus $(m+n)^{2}=m^{2}+n^{2}+2 m n=1+\frac{2}{3}=\frac{5}{3} \Rightarrow m+n=\frac{\sqrt{15}}{3}$. Therefore, $\sin \frac{A-B}{2}+\sin \frac{C}{2}=\sin \frac{A-B}{2}+\cos \frac{A+B}{2}=m y-n x+n y-m x$ $=(m+n)(y-x)=\frac{\sqrt{15}}{3} \times \frac{\sqrt{6}}{3}=\frac{\sqrt{10}}{3}$.