= Solution
Use an expansion for the <eigenstate> continuing the nondegenerate level, with intermediate <normalization> $\langle\psi_0,\psi(\epsilon)\rangle=1$:
$$
E(\epsilon)=E_0+\epsilon E^{(1)}+\epsilon^2E^{(2)}+O(\epsilon^3),\qquad
\psi(\epsilon)=\psi_0+\epsilon\chi+\epsilon^2\eta+\cdots,\qquad\langle\psi_0,\chi\rangle=0.
$$
Insert these expressions into the <eigenvalue> equation. At first order,
$$
(H_0-E_0)\chi=-(H'-E^{(1)})\psi_0.
$$
Taking the <inner product> with $\psi_0$ eliminates the left side by self-adjointness. Thus
$$
\boxed{E^{(1)}=\langle\psi_0,H'\psi_0\rangle,\qquad
E(\epsilon)=\langle\psi_0,H\psi_0\rangle+O(\epsilon^2).}
$$
For $j\ne0$, projection onto $\psi_j$ instead gives
$$
\langle\psi_j,\chi\rangle=\frac{\langle\psi_j,H'\psi_0\rangle}{E_0-E_j}.
$$
At second order, projecting the <eigenvalue> equation onto $\psi_0$ gives $E^{(2)}=\langle\psi_0,H'\chi\rangle$. Expand $\chi$ in the orthonormal energy <basis> and use self-adjointness of $H'$:
$$
\boxed{\epsilon^2E^{(2)}=\epsilon^2\sum_{j\ne0}\frac{|\langle\psi_j,H'\psi_0\rangle|^2}{E_0-E_j}.}
$$
Nondegeneracy ensures the denominators for this level do not vanish; convergence and validity of the expansion are among the assumed perturbation-theory hypotheses.
For the <cosine perturbation of the ground level in a one-dimensional box>, write $u=\pi x/a$. The product-to-sum identity gives
$$
\cos(2u)\sin u=\tfrac12\bigl(\sin(3u)-\sin u\bigr),
$$
so, with the given common <normalization> of the sine eigenfunctions,
$$
\boxed{V\phi_1=\frac\epsilon2(\phi_3-\phi_1).}
$$
The diagonal <matrix> element is $-\epsilon/2$, and the only nonzero off-diagonal <matrix> element out of the <ground state> is $\langle\phi_3,V\phi_1\rangle=\epsilon/2$. Let $E_{\mathrm{box}}=\pi^2\hbar^2/(2ma^2)$. Since $\mathcal E_3=9E_{\mathrm{box}}$, the ground energy is
$$
E_{\mathrm{ground}}=E_{\mathrm{box}}-\frac\epsilon2+\frac{\epsilon^2/4}{E_{\mathrm{box}}-9E_{\mathrm{box}}}+O(\epsilon^3)
=\boxed{\frac{\pi^2\hbar^2}{2ma^2}-\frac\epsilon2-\frac{ma^2\epsilon^2}{16\pi^2\hbar^2}+O(\epsilon^3).}
$$
The second-order term is negative, as expected for a nondegenerate ground level: every excited-<quantum state> denominator is negative. The first-order shift is also negative for positive $\epsilon$, because the ground-<quantum state> density is largest near the center, where this perturbing potential is negative.
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