(1)Q(1,13/4)到拋物線C1的準(zhǔn)線:y=-p/2的距離是13/4+p/2=7/2,p=1/2,
設(shè)拋物線C1：x^2=y上的動點(diǎn)P(t,t^2),
過P作圓C2：x^2+(y-3)^2=1(改題了)的切線：y-t^2=k(x-t),即kx-y+t^2-kt=0,
圓心C2(0,3)到切線的距離=|-3+t^2-kt|/√(k^2+1)=1,
平方得(-3+t^2-kt)^2=k^2+1,
整理得(t^2-1)k^2-2t(t^2-3)k+(t^2-3)^2-1=0,①
若兩切線的斜率之積k1k2=[(t^2-3)^2-1]/(t^2-1)=-1,
則(t^2-3)^2-1=1-t^2,
整理得t^4-5t^2+7=0,無實(shí)數(shù)解,
∴直線PA與PB不垂直.
（2）切線交y軸于A(0,t^2-k1t),B(0,t^2-k2t),
由①,k1+k2=2t(t^2-3)/(t^2-1),
∴AB的中點(diǎn)M:xM=0,yM=t^2-t(k1+k2)/2=t^2-t^2(t^2-3)/(t^2-1)=2t^2/(t^2-1),
由線段AB被直線PQ平分得P,M,Q共線,
PQ,MQ的斜率相等,即(t^2-13/4)/(t-1)=13/4-2t^2/(t^2-1),
兩邊都乘以4(t^2-1),得(4t^2-13)(t+1)=13(t^2-1)-8t^2,
4t^3+4t^2-13t-13
-5t^2 +13=0,
4t^3-t^2-13t=0,解得t1=0,t2,3=(1土√209)/8,
∴點(diǎn)P的坐標(biāo)是(0,0),((1+√209)/8,(105+√209)/32),((1-√209)/8,(105-√209)/32).