= Solution
If the primed frame moves at speed $v$ in the positive $x$ direction relative to the unprimed frame, the one-dimensional <Lorentz transformation> is
$$
\boxed{x'=\gamma_v(x-vt),\qquad t'=\gamma_v(t-vx/c^2),\qquad \gamma_v=(1-v^2/c^2)^{-1/2}.}
$$
For a trajectory with velocity $q=dx/dt$, differentiating these expressions gives $q'=(q-v)/(1-vq/c^2)$. Solving for $q$, when the particle has primed-frame speed $u$, proves the <relativistic velocity-addition formula>
$$
\boxed{q=\frac{u+v}{1+uv/c^2}.}
$$
For the train problem the sign of the frame velocity reverses: the train frame $S$ moves at $+v$ relative to the station frame $S'$. Put $\gamma=(1-v^2/c^2)^{-1/2}$. The transformation appropriate to the subsequent parts is therefore
$$
x'=\gamma(x+vt),\qquad t'=\gamma(t+vx/c^2),\qquad x=\gamma(x'-vt'),\qquad t=\gamma(t'-vx'/c^2).
$$
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