$a|x-1,8|+3,2=4,2$
$⇔|x+1,8|=1$
\(⇔\left[ \begin{array}{l}x+1,8=1\\x+1,8=-1\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=-0,8\\x=-2,8\end{array} \right.\)
$b,(\frac{1}{3})^3.x=$ $(\frac{1}{2})^5$
$⇔\frac{1}{27}x=$ $\frac{1}{32}$
$⇔x=$ $\frac{27}{32}$