= Solution
At threshold, the <pion> is at rest in the final <centre-of-momentum frame>, which moves in the laboratory with
$$
\beta=\frac{E_\gamma^{\rm th}}{m_pc^2+E_\gamma^{\rm th}}=0.1336592\ldots,\qquad
\gamma=\frac{m_pc^2+E_\gamma^{\rm th}}{(m_p+m_\pi)c^2}=1.0090539\ldots.
$$
In the <pion>'s <rest frame>, its <two-photon decay> produces opposite <photons> of equal <energy> $m_\pi c^2/2$. The specified forward and backward directions remain collinear under this boost, so
$$
E_\pm=\frac{m_\pi c^2}{2}\gamma(1\pm\beta).
$$
At threshold, $\gamma(1+\beta)=(m_p+m_\pi)/m_p$ and $\gamma(1-\beta)=m_p/(m_p+m_\pi)$. Hence
$$
\boxed{E_+=77.2148\,\mathrm{MeV},\qquad E_-=59.0075\,\mathrm{MeV}.}
$$
The forward <photon> is the more energetic one. Their product remains $(m_\pi c^2/2)^2$, as required by the invariant <mass> of an oppositely directed <photon> pair, while their sum equals the laboratory <pion> <energy>.
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