$解：(1)x^2-8x+4=x^2-8x+16-16+4=(x-4)^2-12$
$x^2-8x+4=(x-2)^2+4x-8x=(x-2)^2-4x$
$(2)∵x^2+\frac 54y^2+xy-4y+4=0，∴x^2+xy+\frac 14y^2+(y^2-4y+4)=0$
$∴(x+\frac 12y)^2+(y-2)^2=0$
$∴\begin{cases}{x+\frac 12y=0}\\{y-2=0}\end{cases}，解得\begin{cases}{x=-1}\\{y=2}\end{cases}$
$∴x^y=(-1)^2=1$
