$\newcommand{\ket}[1]{\left| #1 \right>}$ $\newcommand{\ad}[0]{\hat{a}^\dagger}$ $\newcommand{\ao}[0]{\hat{a}}$
Firs of all I'd like to tell you that the position operator $x$ is given in terms of the ladder operator by the following relation:
$$\hat x = d (\ao+\ad )$$, where $d=\sqrt{\frac{\hbar}{2m \omega}}$.
Computing $\hat{x}^3$ gives then, as you can easily verify by expanding out $(\ao+\ad)\cdot(\ao+\ad )\cdot (\ao+\ad )$ the following:
$$\hat x^3 = d^3(\ao \ao \ao +\ao \ao \ad + \ao \ad \ao + \ao \ad \ad + \ad \ao \ao + \ad \ao \ad + \ad \ad \ao + \ad \ad \ad)$$
Since I found the notation $:H:$ to be confusing instead I would like to write in calligraphic notation ie I define $\mathcal{H} \equiv \; :H:$ .
$\Delta= \mathcal{H}_{an-harm}-H_{an-harm}$ is then fairly easy to calculate. Notice that $\mathcal{X}^3$ is given by binomial theorem since you don't really care in which order you write the operators as long as $\ad$ is on the far left side. Notice also that $\mathcal{H}_{harm}=H_{harm}$ then it follows that
$$\Delta = \eta d^3(-\ao \ao \ad - \ao \ad \ao - \ao \ad \ad + 2 \ad \ao \ao - \ad \ao \ad + 2 \ad \ad \ao ) $$
You may want to tidy up $\Delta$ by using the commutator identities if you want to have really good answer but I don't find it too necessary to do it for this particular problem.
For the second part you can calculate $\Delta \ket{0}$ and $\mathcal{H}_{an-harm} \ket {0}$ if you remember that $\ao \ket{0} = 0$ and $\ket{n}= \frac{1}{\sqrt{n!} } (\ad)^n \ket{0}$. I'll leave this part to you.