tanx in the question part (i) should be tana.
Equations of motion for the speed of projection, u, in terms of time, t, and projection angle, a.
Vertical: y=utsin(a)-gt^2/2=t(usin(a)-gt/2); horizontal: x=utcos(a).
y/x=(usin(a)-gt/2)/ucos(a)=tan(a)-gt/2ucos(a).
y'=dy/dt=usin(a)-gt; x'=dx/dt=ucos(a).
y'/x'=(usin(a)-gt)/ucos(a)=tan(a)-gt/ucos(a).
(i)
When y'=0, parabolic trajectory is horizontal and gt=usin(a), so t=usin(a)/g at the highest point (vertex).
At t=usin(a)/g, tan(B)=y/x=(usin(a)-usin(a)/2)/ucos(a)=tan(a)/2.
Therefore tan(a)=2tan(B).
(ii)
Slope of normal to the plane has to be -1/tan(B)=-cot(B).
y'/x'=-cot(B)=tan(a)-gt/ucos(a).
gt=(sin(a)/cos(a)+cos(B)/sin(B))ucos(a)=u(sin(a)sin(B)+cos(a)cos(B))/sin(B)=ucos(a-B)/sin(B).
y/x=tan(B) at this time: tan(B)=tan(a)-cos(a-B)/2cos(a)sin(B).
cos(a-B)/2cos(a)sin(B)=tan(a)-tan(B)=(sin(a)cos(B)-sin(B)cos(a))/cos(a)cos(B)=sin(a-B)/cos(a)cos(B).
So, cos(a-B)/2sin(B)=sin(a-B)/cos(B); cos(B)cos(a-B)=2sin(B)sin(a-B) and 2tan(B)tan(a-B)=1.