$解:(1)∵拋物線y=-x2+bx(b為常數)的頂點的橫坐標為-\frac{2×(-1)}=\frac{2},$
$拋物線y=-x2+2x的頂點的橫坐標為-\frac{2}{2×(-1)}=1，$
$∴\frac{2}=1+1，$
$解得b=4$
$(2)∵點A(x_{1},y_{1})在拋物線y=-x2+2x上，點B(x_{1}+t,y_{1}+h)在拋物線y=-x2+4x上，$
$∴y_{1}=-x_{1}2+2x_{1},y_{1}+h=-(x_{1}+t)2+4(x_{1}+t).$
$∴-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\ $
$②將x_{1}=t-1代入h=-t2-2x_{1}t+2x_{1}+4t,得h=-t2-2t(t-1)+2(t-1)+4t，$
$即h=-3t2+8t-2=-3(2-\frac{4}{3})2+\frac{10}{3}$
$∵-3＜0,$
$∴當t=\frac{4}{3}，即x_{1}=\frac{1}{3}時，h取最大值，為\frac{10}{3}$
