I think all three have essentially the same answer.
Consider the static case: E is radially outward from the charge.
Back to the moving case: v/c < 1, so 1 > 1  (v/c)^2 > 0. So in the forward direction (theta = 0) the field is proportional to 1  (v/c)^2 so it is reduced by that factor from the static case. ie it is smaller. In the perpendicular direction theta = pi / 2, and the field is proportional to the inverse square of 1  (v/c)^2, which is greater than 1. This means the field is larger than in the static case. So the field is smaller in the front (and back) and larger out to the sides than it is for a radially symmetric field. The shape is somewhat reminiscent of a flattened disk, ie a "pancake."
1), R is vector from the point charge at PRESENT time, to the field point. This vector is not necessary perpendicular to the velocity vector no matter what velocity, this mean \theta a variable.
Can anyone explain to me?
Thanks
Alan

No idea about this one.
Dan