$解:(1)由(a-2)²+ \sqrt{b-2a}=0,\ $
$得\begin{cases}{ a-2=0, }\ \\ { b-2a=0, } \end{cases}\ 解得\begin{cases}{a=2, }\ \\ {b=4.\ } \end{cases}\ $
$∴b-a的算术平方根是 \sqrt{4-2}=\sqrt{2}.$
$(2)由题意,得\begin{cases}{x-1≥0,\ }\ \\ { 1-x≥0, } \end{cases}\ 解得x=1.\ $
$当x=1时,y=4.$
$∴ \sqrt{x²y}= \sqrt{1²×4}=2.$