7. In $\triangle A B C$, the internal angles $A, B, C$ are opposite the sides $a, b, c$ respectively, and $\sin C \cos \frac{A}{2}=(2-\cos C) \sin \frac{A}{2}$, $\cos A=\frac{3}{5}, a=4$, then the area of $\triangle A B C$ is $\qquad$ .
Solve:
$\cos A=\frac{1-\tan ^{2} \frac{A}{2}}{1+\tan ^{2} \frac{A}{2}}=\frac{3}{5} \Rightarrow \tan \frac{A}{2}=\frac{1}{2}$, thus $\frac{\sin C}{2-\cos C}=\tan \frac{A}{2}=\frac{1}{2}$
$\Rightarrow 2 \sin C+\cos C=2 \Rightarrow 3 \cos ^{2} C-4 \sin C \cos C=0 \Rightarrow C=90^{\circ}$ or $\tan C=\frac{3}{4}$.
When $C=90^{\circ}$, $a=4, b=3, c=5 \Rightarrow S_{\triangle A B C}=6$;
When $\tan C=\frac{3}{4}$, $\sin C=\cos A=\sin \left(90^{\circ}-A\right) \Rightarrow B=90^{\circ}$,
thus $a=4, b=5, c=4 \Rightarrow S_{\triangle A B C}=6$. In summary, $S_{\triangle A B C}=6$.