1题：抛物线符合y²=2px标准型,p=2,焦点（1,0）,渐近线x=-1.
姑且就把直线与x轴的夹角看做倾斜角吧,设倾斜角为α,弦为AB,
AB直线斜率为k,k=tanα,AB过焦点,则AB：y=k(x-1).
设A(x1,y1),B(x2,y2),代入抛物线
y1²=4x1
y2²=4x2,两式相减整理
(y1+y2)(y1-y2)=4(x1-x2)
(y1-y2)/(x1-x2)=4/(y1+y2)
k=4/(y1+y2)
k=4/[k(x1-1)+k(x2-1)]
k=4/[k(x1+x2-2)]
k²=4/(x1+x2-2)
AB弦长=x1+x2+p=x1+x2+2=16/3
x1+x2-2=4/3
k²=3,k=正负根号3
tanα=正负根号3
α=60°或120°
2题：设PQ斜率为k,由题y0≠0知k一定存在.
设P(x1,y1),Q(x2,y2),代入抛物线y²=2px
y1²=2px1
y2²=2px2,两式相减整理
(y1+y2)(y1-y2)=2p(x1-x2)
(y1-y2)/(x1-x2)=2p/(y1+y2)
k=2p/(y1+y2),中点公式得y1+y2=2y0
k=p/y0
3题：设弦为AB,A(x1,y1),B(x2,y2),代入抛物线
y1²=-8x1
y2²=-8x2,两式相减整理
(y1+y2)(y1-y2)=-8(x1-x2)
(y1-y2)/(x1-x2)=-8/(y1+y2)
k=-8/(y1+y2),中点公式得y1+y2=2
k=-4
AB方程为：y-1=-4[x-(-1)]
即y=-4x-3
累死我了,还没有悬赏,你不追加我一点啊?