= Solution
For $E\to\infty$, the outer part of the <action integral for a linear-to-square-root potential> dominates:
$$
f(E)
=\frac{16}{15}(E-1)^{5/2}
+O(E^{3/2})
\sim\frac{16}{15}E^{5/2}.
$$
The <WKB quantization condition> therefore gives
$$
\frac{16}{15}E_n^{5/2}
\sim\pi\varepsilon n.
$$
Taking the power $2/5$ yields the <asymptotic equivalence>
$$
\boxed{
E_n\sim
\left(\frac{15\pi\varepsilon}{16}\right)^{2/5}n^{2/5}.}
$$
Hence
$$
\boxed{
c=\left(\frac{15\pi\varepsilon}{16}\right)^{2/5},
\qquad \alpha=\frac25.}
$$
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