(1)把x=0,y=0代入y=x2+bx+c,得c=0,
再把x=t,y=0代入y=x2+bx,得t2+bt=0,
∵t>0,
∴b=-t;
(2)①不变.
当x=1时,y=1-t,故M(1,1-t),
∵tan∠AMP=1,
∴∠AMP=45°;
②S=S四边形AMNP-S△PAM=S△DPN+S梯形NDAM-S△PAM
= 1/2(t-4)(4t-16)+ 1/2[(4t-16)+(t-1)]×3- 1/2(t-1)(t-1)
= 3/2t²- 15/2t+6.
解 3/2t2- 15/2t+6= 21/8,
得:t1= 1/2,t2= 9/2,
∵4<t<5,
∴t1= 1/2舍去,
∴t= 9/2.
(3) 7/2<t< 11/3.