（1）不等式$2|x|+|x-1|<2$的解集是

（1）不等式$2|x|+|x-1|<2$的解集是

（1）由不等式$2|x|+|x-1|<2$可得

$①\left\{\begin{array}{l}{x<0} \\ {-3 x+1<2}\end{array}\right.$，或$②\left\{\begin{array}{l}{0 \leqslant x<1} \\ {x+1<2}\end{array}\right.$，或$③\left\{\begin{array}{l}{x \geqslant 1} \\ {3 x-1<2}\end{array}\right.$

解$①$得$-\frac{1}{3}故不等式的解集为$\left(-\frac{1}{3},1\right)$故答案为：$\left(-\frac{1}{3},1\right)$（2）点$\left(2\sqrt{2},\frac{\pi}{4}\right)$的直角坐标为$(2,2)$，圆$\rho=4\sin\theta$即$\rho^2=4\rho\sin\theta$转化可得$x^2+(y-2)^2=4$，表示以$(0,2)$为圆心、以$2$为半径的圆由题意可得，圆的切线斜率不存在，故切线方程为$x=2$，化为极坐标方程为$\rho\cos\theta=2$故答案为：$\rho\cos\theta=2$

故不等式的解集为$\left(-\frac{1}{3},1\right)$

故答案为：$\left(-\frac{1}{3},1\right)$

（2）点$\left(2\sqrt{2},\frac{\pi}{4}\right)$的直角坐标为$(2,2)$，圆$\rho=4\sin\theta$即$\rho^2=4\rho\sin\theta$

转化可得$x^2+(y-2)^2=4$，表示以$(0,2)$为圆心、以$2$为半径的圆

由题意可得，圆的切线斜率不存在，故切线方程为$x=2$，化为极坐标方程为$\rho\cos\theta=2$

故答案为：$\rho\cos\theta=2$