= Solution
By the evenness of $V$, the <semiclassical action integral> is twice its integral over the positive half-axis. If $0\leq E\leq1$, then
$$
\begin{aligned}
f(E)
&=2\int_0^{4E}\sqrt{E-\frac x4}\,dx\\
&=8\int_0^E u^{1/2}\,du
=\boxed{\frac{16}{3}E^{3/2}}.
\end{aligned}
$$
If $E\geq1$, split the integral at $x=4$:
$$
f(E)=2\left[
\int_0^4\sqrt{E-\frac x4}\,dx
+\int_4^{(E+1)^2}\sqrt{E+1-\sqrt x}\,dx
\right].
$$
The first integral is
$$
\frac83\left(E^{3/2}-(E-1)^{3/2}\right).
$$
In the second, put $r=\sqrt x$ and then $u=E+1-r$. It becomes
$$
\frac43(E+1)(E-1)^{3/2}
-\frac45(E-1)^{5/2}.
$$
After simplification,
$$
\boxed{
f(E)=
\begin{cases}
\dfrac{16}{3}E^{3/2},&0\leq E\leq1,\\[5pt]
\dfrac{16}{3}E^{3/2}
+\dfrac{16}{15}(E-1)^{5/2},&E\geq1.
\end{cases}}
$$
This is the <action integral for a linear-to-square-root potential>.
The graph begins at the origin, is continuous and strictly increasing, passes through $(1,16/3)$, and tends to infinity. Therefore for every $n\geq0$ and $\varepsilon>0$, the equation
$$
f(E)=\pi\varepsilon\left(n+\frac12\right)
$$
has exactly one solution $E=E_n$ by the <intermediate value theorem> and strict monotonicity.
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