= Solution
Inside the box, the <time-independent Schrödinger equation> is $-\hbar^2\Psi''/(2m)=E\Psi$. The zero endpoint values eliminate the cosine term and require $ka=n\pi$. There is no nonzero zero-<energy> solution with both endpoints zero. <Normalization> uses $\int_0^a\sin^2(n\pi x/a)\,dx=a/2$, so
$$
\boxed{E_n=\frac{n^2\pi^2\hbar^2}{2ma^2},\qquad
\Psi_n(x)=\sqrt{\frac2a}\sin\frac{n\pi x}{a},\qquad n=1,2,\ldots.}
$$
Insert the expansion into $i\hbar\partial_t\psi=H\psi$ and take its <inner product> with $\Psi_m$. <Orthonormality> gives $i\hbar\dot c_m=E_mc_m$ for each $m$, and conversely these equations make every expansion coefficient of the residual zero. Thus $c_n(t)=c_n(0)e^{-iE_nt/\hbar}$ in the usual Hilbert-space or strong-solution class appropriate to the expansion.
For the specified state,
$$
\boxed{\psi(t,x)=\frac35e^{-iE_3t/\hbar}\Psi_3(x)+\frac45e^{-iE_4t/\hbar}\Psi_4(x).}
$$
The modes are real, so the <interference> term is a cosine of the <energy> difference. Writing $s_n=\sin(n\pi x/a)$,
$$
\boxed{|\psi(t,x)|^2=\frac2{25a}\left[9s_3^2+16s_4^2+24s_3s_4\cos\!\left(\frac{7\pi^2\hbar t}{2ma^2}\right)\right].}
$$
Its integral remains one because the cross term integrates to zero. The probabilities of the two <energy> outcomes are fixed at $9/25$ and $16/25$, even though the position <probability density> oscillates.
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