A ball strikes a smooth horizontal floor obliquely and rebounds inelastically. 

Let the ball strike the floor at angle $\mathrm{\alpha }$ with the vertical and velocity at the instant of striking the floor be u and let the ball rebound with velocity v at angle $\mathrm{\beta }$ with the vertical as shown in figure.

Since the floor is smooth, therefore horizontal component of velocity remains unchanged. Hence 

            $\mathrm{v} \sin \mathrm{\beta }=\mathrm{u} \sin \mathrm{\alpha } ...\left(\mathrm{i}\right)$

Since, the floor is inelastic, therefore normal component of velocity just after the collision is less than that just before the collision. Hence,

            $\mathrm{v} \cos \mathrm{\beta }<\mathrm{u} \cos \mathrm{\alpha } ...\left(\mathrm{ii}\right)$

Dividing Eq. (i) by Eq. (ii), $\tan \mathrm{\beta }>\tan \mathrm{\alpha }\text{ or }\mathrm{\beta }>\mathrm{\alpha }$ 

Hence, option (3) is correct.

Since the ball has a horizontal component of velocity, therefore, at highest point, its kinetic energy is not equal to zero.

It means, at highest point, potential energy of the ball is less than the kinetic energy just after rebound. Hence, option (1) is incorrect.

Since the floor is inelastic, therefore there is some loss of energy during the collision. Hence, the total energy of the ball does not remain conserved. So, option (2) is also correct.