= Solution
For $m>0$, $E=\gamma_u mc^2$ and $\mathbf p=\gamma_um\mathbf u$, with $\gamma_u=(1-u^2/c^2)^{-1/2}$. Thus
$$
E^2-p^2c^2=\gamma_u^2m^2c^4(1-u^2/c^2)=m^2c^4.
$$
A <photon> obeys \b[$E=pc$]; $E=hf$ and $p=h/\lambda$ agree because $f\lambda=c$ in vacuum.
Take the incident <photon> along $+x$, and write $\beta=v/c$ for the boost. In the boosted frame the initial momenta are
$$
p'_{\gamma x}=mc\sqrt{\frac{1-\beta}{1+\beta}}=\gamma_vmc(1-\beta),\qquad
p'_{ex}=-\gamma_vm\beta c.
$$
Their sum vanishes when $1-2\beta=0$, giving \b[$v_{\rm ZM}=c/2$] and $\gamma_v=2/\sqrt3$. The initial <photon energy> there is $mc^2/\sqrt3$, and total <energy> is $\sqrt3mc^2$.
For <right-angle Compton scattering in the centre-of-momentum frame>, choose the outgoing ZM <photon> along $+y$. The <electron> has the opposite <momentum> $-p'\widehat{\mathbf y}$. <Conservation of energy> determines its magnitude:
$$
p'c+\sqrt{m^2c^4+p'^2c^2}=\sqrt3mc^2
\quad\Longrightarrow\quad
p'=\frac{mc}{\sqrt3},\qquad E'_e=\frac{2mc^2}{\sqrt3}.
$$
Thus \b[$\mathbf u'_e=-(c/2)\widehat{\mathbf y}$]. Inverting the <velocity> transformation gives
$$
u_{ex}=\frac{u'_{ex}+v}{1+u'_{ex}v/c^2}=\frac c2,\qquad
u_{ey}=\frac{u'_{ey}}{\gamma_v(1+u'_{ex}v/c^2)}=-\frac{\sqrt3c}{4}.
$$
Consequently
$$
\boxed{u_e=\frac{\sqrt7}{4}c=0.6614c,\qquad
\theta_e=-\tan^{-1}\!\left(\frac{\sqrt3}{2}\right)=-40.9^\circ.}
$$
The <electron> recoils below the incident direction for our choice of <photon> scattering above it; the reflected configuration reverses both transverse signs. Finally,
$$
\boxed{E_{\gamma,\rm lab}=\gamma_v(E'_\gamma+vp'_{\gamma x})=\frac23mc^2\simeq341\,\mathrm{keV}.}
$$
As a check, $E_{e,\rm lab}=4mc^2/3$, so the final energies sum to the initial $2mc^2$. The outgoing laboratory <photon> makes $60^\circ$ with $+x$.
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