= Solution
Expand the initial <wavefunction> in the complete orthonormal <energy eigenstates>, with $c_n=\langle\psi_n,\Psi(0)\rangle$. The <unitary time evolution> is
$$
\boxed{\Psi(x,t)=\sum_{n=1}^\infty c_ne^{-iE_nt/\hbar}\psi_n(x),\qquad\sum_{n=1}^\infty|c_n|^2=1.}
$$
On the two-dimensional subspace spanned by $\psi_1,\psi_2$, the <observable> has <matrix> $\begin{pmatrix}2&-1\\-1&2\end{pmatrix}$. Its normalized <eigenvectors> are the symmetric and antisymmetric combinations. A complete orthonormal <eigenbasis> is therefore
$$
\boxed{\phi_1=\frac{\psi_1+\psi_2}{\sqrt2},\quad A\phi_1=\phi_1;\qquad
\phi_2=\frac{\psi_1-\psi_2}{\sqrt2},\quad A\phi_2=3\phi_2;\qquad
\phi_n=\psi_n,\quad A\phi_n=0\ (n\geq3).}
$$
The <eigenvalues> are $1,3,0$, with $0$ infinitely degenerate. The first two are nondegenerate, regardless of the distinct energies $E_n$.
The first <quantum measurement>, with outcome $3$, leaves the state $\phi_2$ up to an irrelevant phase. At time $t$ it has become
$$
\Psi(t)=\frac{e^{-iE_1t/\hbar}\psi_1-e^{-iE_2t/\hbar}\psi_2}{\sqrt2}.
$$
Its <probability amplitude> for outcome $1$ is
$$
\langle\phi_1,\Psi(t)\rangle=\frac12\left(e^{-iE_1t/\hbar}-e^{-iE_2t/\hbar}\right).
$$
The <Born rule> gives the <two-level observable transition probability>
$$
\boxed{\mathbb P(A(t)=1\mid A(0)=3)=\sin^2\left(\frac{(E_2-E_1)t}{2\hbar}\right).}
$$
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