= Solution
First justify the dependence structure rather than assume that adjacent squares behave like an ordinary lag-one ARCH model. Write $Z_t=X_t^2$. Iterating its nonnegative recurrence gives
$$
Z_t=\alpha_0\sum_{k=0}^{m-1}\alpha_2^k\prod_{j=0}^k\varepsilon_{t-2j}^2
+\alpha_2^m\left(\prod_{j=0}^{m-1}\varepsilon_{t-2j}^2\right)Z_{t-2m}.
$$
The product coefficient has <expected value> $\alpha_2^m$, so it tends to zero in probability. The stationary $Z_{t-2m}$ are a tight family; consequently the remainder tends to zero in probability, without needing <independence> between that remainder's factors. The increasing partial sums therefore give the representation
$$
Z_t=\alpha_0\sum_{k\geq0}\alpha_2^k\prod_{j=0}^k\varepsilon_{t-2j}^2\quad\text{almost surely}.
$$
This proves the <parity decomposition of a lag-two ARCH process>: even and odd observations are functions of disjoint noise families. It also shows that the positive scale $\sigma_t$ is a function of past noise, independent of $\varepsilon_t$. The <conditional expectation> of $X_t$ given past noise is zero, and the second-moment recursion is $m_2=\alpha_0+\alpha_2m_2$. <independence> of the parity families gives the adjacent-square <covariance>. Thus
$$
\boxed{\mathbb EX_t=0,\qquad m_2=\mathbb EX_t^2=\frac{\alpha_0}{1-\alpha_2},\qquad
\operatorname{Cov}(X_t^2,X_{t+1}^2)=0.}
$$
To establish finiteness of the <fourth moment> before using its recursion, note that the $L^2$ norm of the $k$th term in the positive series for $Z_t$ is $\alpha_0\sqrt3(\alpha_2\sqrt3)^k$. The <triangle inequality> in <L2 space> makes the series square-integrable when $3\alpha_2^2<1$. <independence> of the current noise and past scale then yields
$$
m_4=3\mathbb E(\alpha_0+\alpha_2X_{t-2}^2)^2
=3\alpha_0^2+6\alpha_0\alpha_2m_2+3\alpha_2^2m_4,
$$
and hence
$$
\boxed{\mathbb EX_t^4=\frac{3\alpha_0^2(1+\alpha_2)}{(1-\alpha_2)(1-3\alpha_2^2)}\quad\text{if }3\alpha_2^2<1.}
$$
If $3\alpha_2^2\geq1$, a finite $m_4$ would make $(1-3\alpha_2^2)m_4=3\alpha_0^2+6\alpha_0\alpha_2m_2>0$, which is impossible. Its <fourth moment> is then infinite. The adjacent-square product remains integrable by parity <independence>, even in that case.
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