Given $\cos \left(\frac{\pi}{4}+x\right)=\frac{3}{5}, \frac{17 \pi}{12}<x<\frac{7 \pi}{4}$, find $\frac{\sin 2 x+2 \sin ^{2} x}{1-\tan x}$.
$$\begin{array}{l}
\text { Method 1: } \because \cos \left(\frac{\pi}{4}+x\right)=\frac{3}{5}, \quad \therefore \cos x \\
-\sin x=\frac{3}{5} \sqrt{2}, \quad 1-2 \sin x \cos x=\frac{18}{25}, \quad(\sin x \\
+\cos x)^{2}=1+2 \sin x \cos x=\frac{32}{25} \text {, and } \frac{17 \pi}{12}<x \\
<\frac{7 \pi}{4} \text {, so } \sin x+\cos x<0, \therefore \sin x+\cos x= \\
-\frac{4 \sqrt{2}}{5} \text {, solving the system of equations }\left\{\begin{array}{l}
\cos x-\sin x=\frac{3 \sqrt{2}}{5} \\
\cos x+\sin x=\frac{-4 \sqrt{2}}{5}
\end{array}\right. \text {, } \\
\text { we get }\left\{\begin{array}{l}
\cos x=\frac{-\sqrt{2}}{10} \\
\sin x=\frac{-7 \sqrt{2}}{10}
\end{array}, \tan x=7 . \quad \therefore \text { the original expression }=\right. \\
\frac{2 \sin x \cos x+2 \sin ^{2} x}{1-\tan x}= \\
2 \times\left(-\frac{7 \sqrt{2}}{10}\right) \times\left(-\frac{\sqrt{2}}{10}\right)+2 \times\left(-\frac{7 \sqrt{2}}{10}\right)^{2}
\end{array}$$