$解:(1)∵抛物线y=-x²+bx(b为常数)的顶点的横坐标为-\frac{b}{2×(-1)}=\frac{b}{2},$
$抛物线y=-x²+2x的顶点的横坐标为-\frac{2}{2×(-1)}=1，$
$∴\frac{b}{2}=1+1，$
$解得b=4$
$(2)∵点A(x_{1},y_{1})在抛物线y=-x²+2x上，点B(x_{1}+t,y_{1}+h)在抛物线y=-x²+4x上，$
$∴y_{1}=-x_{1}²+2x_{1},y_{1}+h=-(x_{1}+t)²+4(x_{1}+t).$
$∴-x_1²+2x_{1}+h=-(x_1²+t)²+4(x_{1}+t).整理，得h=-t²-2x_{1}t+2x_{1}+4t\ $
$①∵h=3t，$
$∴3t=-t²-2x_{1}t+2x_{1}+4t，即t(t+2x_{1})=t+2x_{1}，$
$∵x_{1}≥0,t＞0,∴t+2x_{1}＞0.$
$∴t=1.$
$∴h=3\ $
$②将x_{1}=t-1代入h=-t²-2x_{1}t+2x_{1}+4t,得h=-t²-2t(t-1)+2(t-1)+4t，$
$即h=-3t²+8t-2=-3(2-\frac{4}{3})²+\frac{10}{3}$
$∵-3＜0,$
$∴当t=\frac{4}{3}，即x_{1}=\frac{1}{3}时，h取最大值，为\frac{10}{3}$
