?$解:(1)∵拋物線y=x2-2mx+m2+2m-1過點(diǎn)B(3,5)，$?
?$∴把B(3,5)代入y=x2-2mx+m2+2m-1.整理，得m2-4m+3=0，$?
?$解得m=1，m_{2}=3.$?
?$當(dāng)m=1時(shí)，y=x2-2x+2=(x-1)2+1,其頂點(diǎn)A的坐標(biāo)為(1,1);$?
?$當(dāng)m=3時(shí)，y=x2-6x+14=(x-3)2+5,其頂點(diǎn)A的坐標(biāo)為(3,5)$?
?$綜上所述，頂點(diǎn)A的坐標(biāo)為(1,1)或(3,5)$?
?$(2)∵y=x2-2mx+m2+2m-1=(x-m)2+2m-1,$?
?$∴頂點(diǎn)A的坐標(biāo)為(m,2m-1).$?
?$由\begin{cases}{x=m,\ }\\{y=2m-1，}\end{cases}得y=2x-1.$?
?$∴y關(guān)于x的函數(shù)表達(dá)式為y=2x-1$?
