= Solution
The original PDF threshold contains $\hbar$, not $h$. It is $8E_1$, so the energies strictly above it are exactly those with $n\geq3$. By the <Born rule>, the required probability is $1-|c_1|^2-|c_2|^2$. The two excluded amplitudes are
$$
c_1=\frac{2\sqrt3}{\pi}\left(1-\frac2\pi\right),\qquad c_2=-\frac{\sqrt3}{\pi}.
$$
Consequently
$$
\boxed{\mathbb P(E>8E_1)=1-\frac{15}{\pi^2}+\frac{48}{\pi^3}-\frac{48}{\pi^4}.}
$$
Thus, in increasing powers of $1/\pi$,
$$
\boxed{(b_0,b_1,b_2,b_3,b_4)=(1,0,-15,48,-48).}
$$
The threshold lies strictly between $E_2$ and $E_3$, so no equality-at-threshold convention affects the answer.
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