\(a,\Rightarrow\left|x\right|=\dfrac{1}{3}-x\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(ktm\right)\end{matrix}\right.\\ \Rightarrow x=\dfrac{1}{6}\\ b,\Rightarrow\left|x\right|=x+2\Rightarrow\left[{}\begin{matrix}x=x+2\left(x\ge0\right)\\x=-x-2\left(x< 0\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}0x=2\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\ \Rightarrow x=-1\)