= Solution
The <ladder operators> obey $[a,a^\dagger]=1$ and $N=a^\dagger a$ has $[N,a]=-a$, $[N,a^\dagger]=a^\dagger$. Since $\langle\psi,N\psi\rangle=\|a\psi\|^2\geq0$, its <eigenvalues> are nonnegative. If $N|\nu\rangle=\nu|\nu\rangle$, lowering gives <eigenvalue> $\nu-1$ and squared norm $\nu$. Repeated lowering cannot produce a negative nonzero eigenstate; the norm product $\nu(\nu-1)\cdots$ forces $\nu$ to be an integer. The final lowered state has $a|0\rangle=0$. With a unique ground state all normalized states are
$$
|n\rangle=\frac{(a^\dagger)^n}{\sqrt{n!}}|0\rangle,\qquad
\boxed{E_n=\hbar\omega(n+\tfrac12)},\quad n=0,1,\ldots.
$$
For <time-independent perturbation theory>, write $H'=H+\lambda W$ with $W=\hbar\omega(a^2+a^{\dagger2})$. The standard first correction is $\lambda\langle n|W|n\rangle$ and the second is $\lambda^2\sum_{j\ne n}|\langle j|W|n\rangle|^2/(E_n-E_j)$. The first vanishes. Only $j=n-2,n+2$ occur, with ladder coefficients $\sqrt{n(n-1)}$ and $\sqrt{(n+1)(n+2)}$, respectively. Therefore
$$
\Delta E_n^{(2)}=\lambda^2\hbar\omega\left[\frac{n(n-1)}2-\frac{(n+1)(n+2)}2\right]
=-\lambda^2\hbar\omega(2n+1).
$$
The downward term is zero for $n<2$.
Inverting the given position/momentum expressions gives
$$
a^2+a^{\dagger2}=\frac{m\omega}{\hbar}\widehat x^2-\frac{\widehat p^2}{m\hbar\omega}.
$$
Thus \b[$\boxed{\alpha=1-2\lambda,\ \beta=1+2\lambda}$]. For $|\lambda|<1/2$ both are positive, and this is a <quantum harmonic oscillator> of effective mass $m/\alpha$ and frequency $\omega\sqrt{\alpha\beta}$. Its exact energies are
$$
\boxed{E_n'=\hbar\omega\sqrt{1-4\lambda^2}(n+\tfrac12).}
$$
Expansion gives $E_n'=\hbar\omega(n+1/2)[1-2\lambda^2+O(\lambda^4)]$, agreeing with both perturbative corrections.
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