= Solution
Each <energy eigenstate> acquires the phase $e^{-iE_nt/\hbar}$ under <unitary time evolution>. Consequently \b[at the requested time $t=1$],
$$
\boxed{\psi(x,1)=\frac1{\sqrt{10}}\left(3e^{-iE_1/\hbar}\phi_1(x)-e^{-i9E_1/\hbar}\phi_3(x)\right),\qquad 0<x<a,}
$$
and the <wavefunction> is zero outside the well. Equivalently, in elementary functions,
$$
\psi(x,1)=\frac1{\sqrt{5a}}\left(3e^{-i\pi^2\hbar/(2ma^2)}\sin\frac{\pi x}{a}-e^{-i9\pi^2\hbar/(2ma^2)}\sin\frac{3\pi x}{a}\right).
$$
The distinct phases retain their relative sign from the initial expansion; an overall <global phase> would not change measurement probabilities.
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