= Solution
In the two-dimensional span of the first two <energy eigenstates>, the measured <observable> has matrix $S=\begin{pmatrix}7&24\\24&-7\end{pmatrix}$. Its square is $625I$, and its trace is zero, giving <eigenvalues> $25,-25$. The orthogonal complement has eigenvalue zero. Thus the full spectrum is $\boxed{\{-25,0,25\}}$.
The normalized eigenstates for $-25$ and $25$ are respectively
$$
\chi_- =\frac{3\psi_1-4\psi_2}{5},\qquad \chi_+=\frac{4\psi_1+3\psi_2}{5}.
$$
The lowest measurement outcome prepares $\chi_-$ up to an irrelevant overall phase. The subsequent <unitary time evolution> gives
$$
\chi_-(t)=\frac35e^{-iE_1t/\hbar}\psi_1-\frac45e^{-iE_2t/\hbar}\psi_2.
$$
By the <Born rule>, the probability of measuring the lowest eigenvalue again is the squared projection onto its one-dimensional <eigenspace>:
$$
\begin{aligned}
P_-&=|\langle\chi_-,\chi_-(t)\rangle|^2=\left|\frac9{25}e^{-iE_1t/\hbar}+\frac{16}{25}e^{-iE_2t/\hbar}\right|^2\\
&=\boxed{\frac{337+288\cos((E_1-E_2)t/\hbar)}{625}}.
\end{aligned}
$$
This is the <two-level quantum return probability> with $q=9/25$. At $t=0$ the probability is one; if the two energies happen to be equal it remains one for every time.
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