= Solution
In the <position> representation $p=-i\hbar\,d/dx$, so $a\psi_0=0$ becomes
$$
\left(\frac{d}{dx}+\frac{m\omega}{\hbar}x\right)\psi_0=0.
$$
Integrating and using the <Gaussian integral> for <wavefunction normalization> gives
$$
\boxed{\psi_0(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\left(-\frac{m\omega x^2}{2\hbar}\right).}
$$
The <creation operator> gives $|1\rangle=a^\dagger|0\rangle$, and therefore
$$
\boxed{\psi_1(x)=\sqrt{\frac{2m\omega}{\hbar}}\,x\psi_0(x).}
$$
Its <norm> is one since $\int x^2|\psi_0|^2dx=\hbar/(2m\omega)$. Moreover $\psi_0$ is even and $\psi_1$ is odd, so the overlap integrand $\psi_0^*\psi_1$ is odd and integrable. Integration over the whole real line gives $\boxed{\langle0|1\rangle=0}$, establishing <orthogonality>.
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