We apply the work-energy theorem, where the change in mechanical energy equals the work done by the non-conservative force of friction: $\Delta E_{mech} = W_{f}$. Let the initial level at point A be the reference for potential energy ($y_A=0$).

The work done by friction over a distance $s$ on the incline is $W_f = -f_k s = -(\mu_k N) s$. The normal force is $N = mg\cos\theta$, so $W_f = -\mu_k mgs\cos\theta$.

The change in mechanical energy from A to a final point F is $\Delta E_{mech} = (\frac{1}{2}mv_F^2 + mgy_F) - \frac{1}{2}mv_A^2$. Combining these gives:

$$(\frac{1}{2}mv_F^2 + mgy_F) - \frac{1}{2}mv_A^2 = -\mu_k mgs\cos\theta$$

First, we determine if the block reaches point B. This occurs if the final kinetic energy at B is non-negative. At point B, the final height is $y_F = y_B = h + L\sin\theta$ and the distance of friction is $s=L$.

$$\frac{1}{2}mv_B^2 + mg(h + L\sin\theta) - \frac{1}{2}mv_A^2 = -\mu_k mgL\cos\theta$$ $$v_B^2 = v_A^2 - 2g(h + L\sin\theta) - 2g\mu_k L\cos\theta = v_A^2 - 2g(h + L(\sin\theta + \mu_k\cos\theta))$$

The block reaches B if $v_B^2 \ge 0$.

[Q1] If the block reaches B, its speed is found from the expression for $v_B^2$:

$$v_B = \sqrt{v_A^2 - 2g(h + L(\sin\theta + \mu_k\cos\theta))}$$

[Q2] If the block does not reach B ($v_B^2 < 0$), it stops momentarily at a distance $d < L$ along the rough section. At this point, the final speed is $v_F=0$. The maximum height is $H_{max} = h + d\sin\theta$, and the friction distance is $s=d$. The work-energy equation becomes:

$$(0 + mg(h + d\sin\theta)) - \frac{1}{2}mv_A^2 = -\mu_k mgd\cos\theta$$

Solving for $d$:

$$mgd(\sin\theta + \mu_k\cos\theta) = \frac{1}{2}mv_A^2 - mgh$$ $$d = \frac{v_A^2 - 2gh}{2g(\sin\theta + \mu_k\cos\theta)}$$

The maximum height is then:

$$H_{max} = h + d\sin\theta = h + \frac{(v_A^2 - 2gh)\sin\theta}{2g(\sin\theta + \mu_k\cos\theta)}$$