$解:(1)点A(\frac{3}{2},-\frac{1}{2})是“好点”,理由如下:\ $
$当A(\frac{3}{2},-\frac{1}{2})时,m-1=\frac{3}{2},$
$\frac{n+2}{2}=-\frac{1}{2},$
$得m=\frac{5}{2},n=-3$
$则2m=5,8+n=5$
$∴2m=8+n$
$∴点A(\frac{3}{2},-\frac{1}{2})是“好点” ,$
$点B(4,10)不是“好点”$
$理由如下: 当B(4,10)时,m-1=4,$
$\frac{n+2}{2}=10,得m=5,n=18$
$则2m=10,8+n=26$
$∴2m≠8+n$
$∴点B(4,10)不是“好点”$