This problem involves two main stages: the collision and the swing.

1. Perfectly Inelastic Collision: The bullet embeds in the box. Horizontal momentum is conserved. Let $V$ be the velocity of the combined system (mass $M+m$) immediately after the collision.

   $$mv = (M+m)V$$

2. Pendulum Swing: After the collision, the mechanical energy of the combined system is conserved as it swings upwards. The initial kinetic energy is converted into gravitational potential energy at the peak height $h$. The height is related to the angle by $h = L(1-\cos\theta)$.

   $$\frac{1}{2}(M+m)V^2 = (M+m)gh = (M+m)gL(1-\cos\theta)$$ $$V = \sqrt{2gL(1-\cos\theta)}$$

3. Bullet Velocity (Q1): Substitute the expression for $V$ from the swing into the momentum equation.

   $$v = \frac{M+m}{m}V = \frac{M+m}{m}\sqrt{2gL(1-\cos\theta)}$$

4. Energy Loss (Q2): The mechanical energy loss $\Delta E$ occurs during the collision. It is the initial kinetic energy of the bullet minus the kinetic energy of the combined system immediately after the collision.

   $$\Delta E = \frac{1}{2}mv^2 - \frac{1}{2}(M+m)V^2$$

   Using the energy conservation from the swing, $\frac{1}{2}(M+m)V^2 = (M+m)gL(1-\cos\theta)$. Substituting the expression for $v$:

   $$\Delta E = \frac{1}{2}m\left(\frac{M+m}{m}\sqrt{2gL(1-\cos\theta)}\right)^2 - (M+m)gL(1-\cos\theta)$$ $$\Delta E = \frac{(M+m)^2}{m}gL(1-\cos\theta) - (M+m)gL(1-\cos\theta)$$ $$\Delta E = (M+m)gL(1-\cos\theta)\left(\frac{M+m}{m} - 1\right) = \frac{M(M+m)}{m}gL(1-\cos\theta)$$