$解：? (1)?∵四邊形?AFPE?是平行四邊形$
$∴?PF//CA?$
$∴?△BFP∽△BAC?$
$∴?\frac {S_{△BFP}}{S_{△ABC}}=(\frac {x}{2})2?$
$∴?S_{△ABC}=1?$
$∴?S_{△BFP}=\frac {x2}{4}?$
$同理：?S_{△PEC}=(\frac {2-x}{2})2?$
$∴?y=1-\frac {x2}{4}-\frac {4-4x+x2}{4}?$
$∴?y=-\frac {x2}{2}+x?$
$?(2)?上述函數(shù)有最大值，最大值為?\frac {1}{2}；?理由如下$
$?y=-\frac {x2}{2}+x=-\frac {1}{2}(x-1)2+\frac {1}{2}，??-\frac {1}{2}\lt 0?$
$∴?y?有最大值$
$∴當(dāng)?x=1?時(shí)，?y?有最大值，最大值為?\frac {1}{2}?$
