(1) If the mass of the hoop is m, and its radius is r, its moment of inertia I=mr².
The height of the incline is Lsinθ where θ is the incline angle and L the length of the slope.
Because energy is conserved, initial energy=final energy.
Initial energy is mgLsinθ, the potential energy because of gravity.
Final energy, translational kinetic energy plus rotational kinetic energy, at the end of the slope is ½mv²+½I⍵² where v is the final velocity and ⍵ is the angular velocity.
But ⍵=v/r so the rotational kinetic energy is ½(mr²)(v²/r²)=½mv². The total final energy is mv².
Therefore mgLsinθ=mv² and v=√(gLsinθ) because the m’s cancel out.
If L=1 and θ=37°, v=√9.81sin37=2.43m/s approx, the speed of the hoop when it reaches the base of the incline.
(2) There are no horizontal forces acting on the hoop once it rolls off the incline (friction is ignored) so its velocity doesn’t change. If we assume that its (tangential) velocity remains the same and is horizontal the vertical component of the velocity is zero. So the time t taken to drop off the table in free fall is given by 0.80=½gt² then t=√(1.60/g)=0.404 seconds approx. In that time the hoop is displaced horizontally 2.43×0.404=0.98m from the foot of the table.