This is a fun, high-quality qualifying exam question. The algebra is not hard; the physical insight takes some real thought; there are many ways to be partly right. Here's my take on it.
Acceleration
From Einstein's equation $E^2 = p^2 + m^2$ we have for each photon $E = p = hf_0$ (in the reference frame of the laser). We can use the power of the laser $P_0$ to find the rate at which individual photons are emitted:
$$P_0 = \dot N_0 hf_0.$$
(Pardon me for writing $\dot N$ rather than $dN/dt$.) The total force exerted by the laser is the time derivative of its momentum, which is the time derivative of its energy, which is just the power,
$$F = \dot p = \dot E = P_0,$$
and since the laser's momentum must be coming from the rocket we have its acceleration is $a = P_0/m_0$. You'll have to stick some $c$s back in to make the dimensions come out right.
Received Power
If the rocket is moving away from Earth at constant speed $\beta = v/c$, with corresponding relativistic factor $\gamma = 1/\sqrt{1-\beta^2}$, there are three factors that affect the power received at Earth:
- Each photon's frequency is red-shifted to $f = f_0\sqrt{\frac{1-\beta}{1+\beta}}$.
- Time on the rocket will be dilated, reducing the rate at which photons appear to be emitted to $\dot N = \dot N_0 / \gamma$.
- Each photon emitted by the rocket has a little bit further to travel back to Earth. If the time between photon emissions (in Earth's frame) is $\Delta t = 1 / \dot N$, each photon has to go $\Delta x = v \Delta t$ further than the last, adding an extra delay $\Delta x / c$ to its trip. So the interval between photons arriving at Earth is
$$\Delta t' = \Delta t \cdot (1 + \beta) .$$
The rate at which photons reach Earth is therefore $\dot N' = 1 / \Delta t'$.
Combining these we have a power recieved at Earth of
$$P' = \dot N' hf = \frac{P_0}{\gamma(1+\beta)}\sqrt\frac{1-\beta}{1+\beta}
% = \frac{P_0}{\gamma^2(1+\beta)^2}
= P_0 \frac{1-\beta}{1+\beta}
.$$
It's always possible in relativity problems to get identical results using classical EM fields for light instead of photons, with Poynting vectors carrying momentum, etc. I would not know how to go about that in this case.
Final rest mass
This part was not immediately obvious to me. The messy option is to try and integrate the expression from the previous section; that probably requires assumptions about the time profile of the acceleration. Usually when you only know the initial and final conditions for a problem, conservation of energy is a good strategy. I wasted some time before I remembered to use conservation of momentum, too.
We know that the final momentum of the rocket is $p_f = \gamma_f v_f m_f$, and that the combined momentum of the rocket and its laser exhaust, in the initial rest frame, is zero. Then, using Einstein's equation again, we have a bunch of backwards-going photons with total energy
$$E_{\text{exhaust}} = \left|-p_f\right|$$
and a rocket with total energy
$$E_{\text{rocket}}^2 = p_f^2 + m_f ^2.$$
If we assume that none of the laser power was wasted as heat, these components must add up to the initial rest mass of the rocket,
\begin{align}
p_f + \sqrt{p_f^2 + m_f^2} &= m_0
\\
m_f \left( \gamma_f v_f + \sqrt{\gamma_f^2 v_f^2 + 1}
\right) &= m_0
\\
m_f \left( \gamma_f v_f + \gamma_f
\right) &= m_0
\\
m_f &= m_0 \sqrt{\frac{1-v_f}{1+v_f}}
\end{align}
Actually this result also holds if the laser power supply is inefficient, so long as the rocket is thermally insulated such that the thermal waste photons are all emitted in the same direction as the exhaust --- tail of the rocket hot, head of the rocket cold. If there's forward-going thermal radiation, it will be focused in the forward direction in a complicated way, and the problem becomes much more difficult.
For parallel boosts we can treat space as $1$-dimensional, so Lorentz boosts are in $2$ dimensions. They are hyperbolic rotation matrices, of the form $$\left(\begin{array}{rl}
\cosh w & \sinh w\\
\sinh w & \cosh w
\end{array}\right),$$which multiply viz.$$\left(\begin{array}{rl}
\cosh w_{1} & \sinh w_{1}\\
\sinh w_{1} & \cosh w_{1}
\end{array}\right)\left(\begin{array}{rl}
\cosh w_{2} & \sinh w_{2}\\
\sinh w_{2} & \cosh w_{2}
\end{array}\right)=\left(\begin{array}{rl}
\cosh\left(w_{1}+w_{2}\right) & \sinh\left(w_{1}+w_{2}\right)\\
\sinh\left(w_{1}+w_{2}\right) & \cosh\left(w_{1}+w_{2}\right)
\end{array}\right).$$As @Qmechanic's second point notes, you're just adding $w$s.
Best Answer
You cannot just combine the Lorentz boosts in that way - the boosts do not form a subgroup of the Lorentz group, the successive application of two boosts is, in general, not a boost, but a boost followed by a rotation, as you may see by explicitly writing down the $4\times 4$-matrices corresponding to the boosts and computing their product, which simply isn't the matrix of a pure boost. This phenomenon is known as Thomas precession.
Nevertheless, one can derive a formula for the composition of two boosts in terms of the velocities, but the formula for the rotation induced by the two boosts is... either ugly or requires going on about gyrovectors, so I will not reproduce it here.