= Solution
Use the normalized <eigenfunctions> of the <infinite square well>,
$$
\phi_n(x)=\sqrt{\frac2a}\sin\frac{n\pi x}{a},\qquad E_n=\frac{n^2\pi^2\hbar^2}{2ma^2},\quad 0<x<a.
$$
The sine identity expands the initial <wavefunction> as
$$
\psi_0=\frac{3\phi_1-\phi_3}{\sqrt{10}}.
$$
The normalization factor $4/\sqrt{5a}$ yields precisely these coefficients. By the <Born rule>, the <energy measurement> probabilities are $9/10$ for $E_1$ and $1/10$ for $E_3=9E_1$. Thus \b[the energy expectation is]
$$
\boxed{\langle E\rangle=\frac9{10}E_1+\frac1{10}E_3=\frac{9\pi^2\hbar^2}{10ma^2}.}
$$
The two probabilities sum to one, also checking the <wavefunction normalization>.
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