The principle of conservation of mechanical energy applies as the incline is frictionless. Let the release point of the block be the reference level for gravitational potential energy ($U_{g,i} = 0$).

[Q1] The elastic potential energy stored in a spring compressed by a distance $x$ is given by:

$$U_e = \frac{1}{2}kx^2$$

[Q2, Q3] The system's initial energy $E_i$ is entirely the elastic potential energy of the spring, as the block is at rest ($K_i=0$) and at the reference height ($U_{g,i}=0$).

$$E_i = U_e = \frac{1}{2}kx^2$$

At the highest point, a distance $d$ up the incline, the block momentarily comes to rest ($K_f=0$). The block is no longer in contact with the spring, so the final elastic potential energy is zero ($U_{e,f}=0$). The final gravitational potential energy $U_{g,f}$ is due to the vertical height gain $h = d\sin\theta$.

$$E_f = U_{g,f} = mgh = mgd\sin\theta$$

By conservation of energy, $E_i = E_f$:

$$\frac{1}{2}kx^2 = mgd\sin\theta$$

[Q2] The change in gravitational potential energy is $\Delta U_g = U_{g,f} - U_{g,i} = mgd\sin\theta$. From the energy conservation equation, we see that this is equal to the initial elastic potential energy.

$$\Delta U_g = \frac{1}{2}kx^2$$

[Q3] To find the distance $d$, we rearrange the energy conservation equation:

$$d = \frac{kx^2}{2mg\sin\theta}$$