= Solution
Put $u=a+i\omega t$. Then
$$
\dot F=i\omega A\,\operatorname{sech}^2u,
\qquad
\dot G=-i\omega
\sqrt{\frac{m\omega}{\hbar}}\,
\operatorname{sech}u\tanh u.
$$
The $G$ equation from part (b) is satisfied when
$$
\frac{2\hbar^2A}{m}=\hbar\omega,
$$
so
$$
\boxed{A=\frac{m\omega}{2\hbar}}.
$$
With this value, the $F$ equation follows from $\tanh^2u=1-\operatorname{sech}^2u$: both its constant and $\operatorname{sech}^2u$ coefficients agree. Thus the stated $F,G$ satisfy both shape equations; the remaining scalar equation determines the normalization factor $N(t)$.
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