?$解:(1)由題意可得:\begin{cases}{0=c}\\{0=16a+4b+c}\\{3=a+b+c}\end{cases}$?
?$解得a=-1,b=4,c=0$?
?$∴二次函數(shù)的表達(dá)式為y=-x2+4x$?
?$設(shè)直線AB的表達(dá)式為y=kx+n$?
?$∴\begin{cases}{0=4k+n}\\{3=k+n}\end{cases}$?
?$解得k=-1,n=4$?
?$∴直線AB對(duì)應(yīng)的函數(shù)表達(dá)式為y=-x+4$?
?$當(dāng)x=0時(shí),y=4$?
?$∴C(0,4)$?
?$(2)∵點(diǎn)P在直線AB上方，$?
?$∴ 1＜m＜4.$?
?$由題意，得P(m，-m2+4m)、D(m，-m+4).$?
?$∴PD=y_P-y_D=-m2+4m+m-4=-m2+5m-4=-(m-\frac{5}{2})2+\frac{9}{4}$?
?$∵ -1＜0,$?
?$∴ 當(dāng)m=\frac{5}{2}時(shí)，PD的長(zhǎng)取得最大值，最大值為\frac{9}{4}$?
