$解: (1)设y= ax²+ bx+2,把(1,1)(3,5)代入$
$得:\begin{cases}{a+b+2=1}\\{9a+3b+2=5} \end{cases}$
$解得a=1,b=-2$
$所以y= x²- 2x+2$
$(2)设y=a(x+1)²+2,$
$把(2,1)代入得9a+2= 1,$
$得a=-\frac {1}{9}$
$所以y=-\frac {1}{9}x²-\frac {2}{9}x+\frac {17}{9}$
$(3)设y=a(x+ 1)(x- 2),$
$把(1 , 2)代入得-2a=2,得a=-1$
$所以y= -(x+1)(x-2)= -x²+x+2$