(1)由题意知|FQ|=4+
p
2=5,∴p=2.∵m2=2×2×4,∴m=±4
(2)由题意知直线L的斜率存在,设为k,则直线L的方程为:y=k(x-1),代入抛物线方程:y2=4x,得
k2x2-(2k2+4)x+k2=0,设点A(x1,y1),B(x2,y2),∴x1+x2=
2k2+4
k2,x1x2=1;
又∵|AB|=
1+k2|x1-x2|=8,|AB|=
1+12
(x1+x2)2−4x1x2=8∴
1
k4+
1
k2−2=0∴k2=1∴k=±1;
∴所求直线方程为:x-y-1=0或x+y-1=0.