Conditional variance vs. unconditional variance in ARCH model
I cannot directly answer your question, but I think I can shed some light. What I seem to show is that under some restrictions, the unconditional variance is finite. However, I am not sure how to relate my answer to the notion of volatility clustering.

Note that if we assume stationarity, then $\sigma^2=E(\epsilon_t^2)$ for all $t$ and it follows by the law of total expectations that $$\sigma^2=E(E(\epsilon_t^2|I_{t-1}))=E(h_t)=\alpha_0+\alpha_1\sigma^2+\alpha_2\sigma^2$$ using your expression for $h_t$. Thus, $\sigma^2=\frac{\alpha_0}{1-(\alpha_1+\alpha_2)}$ is a solution whenever $\alpha_1+\alpha_2&lt;1$. 

Now, by the law of total variance, \begin{align*}V(Y_t)&amp;=E(V(Y_t|I_{t-1}))+V(E(Y_t|I_{t-1}))\\
&amp;=E(\epsilon_t^2)+V(\mu_t)\\
&amp;=\sigma^2+a_1^2V(Y_{t-1}) \end{align*} using your results, where $\mu_t=a_0+a_1 Y_{t-1}$. Assuming stationarity, and thus that the variance is constant for all $t$, it follows that $$V(Y_t)=\frac{\sigma^2}{1-a_1^2}$$ is a solution for $a_1\neq\pm 1$ and $\alpha_1+\alpha_2&lt;1$.

I encourage you to clearly note my assumptions. I cannot further assess this answer as I know next to nothing about time series. 

(Also note that we from the assumption of stationarity have that $E(Y_t)=a_0+a_1E(Y_{t-1})$ implies that $$E(Y_t)=\frac{a_0}{1-a_1}$$ is a solution.)