= Solution
Write $\hat N=\hat a^\dagger\hat a$. The <creation and annihilation operators> obey
$$
[\hat N,\hat a^\dagger]
=\hat a^\dagger[\hat a,\hat a^\dagger]=\hat a^\dagger,
\qquad
[\hat H,\hat a^\dagger]=\hbar\omega\hat a^\dagger.
$$
Consequently,
$$
\begin{aligned}
\hat H\hat a^\dagger\Psi_0
&=(\hat a^\dagger\hat H+[\hat H,\hat a^\dagger])\Psi_0\\
&=(E_0+\hbar\omega)\hat a^\dagger\Psi_0
=\boxed{\frac32\hbar\omega\,\hat a^\dagger\Psi_0}.
\end{aligned}
$$
The resulting function is not zero: its squared norm is
$$
\|\hat a^\dagger\Psi_0\|^2
=\langle\Psi_0|\hat a\hat a^\dagger|\Psi_0\rangle
=\langle\hat a^\dagger\hat a+1\rangle=1.
$$
Thus \b[$\hat a^\dagger\Psi_0$ is the normalized first excited state], up to phase. More generally the <creation operator> raises the energy by one quantum $\hbar\omega$: $\hat a^\dagger|n\rangle=\sqrt{n+1}\,|n+1\rangle$. The factor is required because applying it to a higher state does not preserve normalization.
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