= Solution
The weight is even, so opposite-parity <polynomials> are automatically orthogonal. Set $p_0=1$, $p_1=x$, then subtract lower-degree components within each parity class. The needed moments are
$$
\mu_0=\frac83,\qquad\mu_2=\frac{16}{15},\qquad\mu_4=\frac{24}{35}.
$$
Thus $p_2=x^2-\mu_2/\mu_0=x^2-2/5$ is orthogonal to $p_0$, and $p_3=x^3-(\mu_4/\mu_2)x=x^3-9x/14$ is orthogonal to $p_1$. Parity supplies every other pair. One valid answer is
$$
\boxed{p_0=1,\quad p_1=x,\quad p_2=x^2-\frac25,\quad p_3=x^3-\frac9{14}x.}
$$
All have the required degree and positive norm; arbitrary nonzero rescalings are equally valid.
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