= Solution
{c}
The weight is strictly positive and all its <polynomial> moments are finite. Thus the given integral is a <inner product>: for a nonzero <polynomial>, its square is positive on a nonempty interval, so its weighted integral is strictly positive.
In the <finite-dimensional vector space> $V_n$, subtract the <orthogonal projection onto a finite-dimensional subspace> of $x^n$ onto $V_{n-1}$:
$$
\phi_n=x^n-\operatorname{proj}_{V_{n-1}}x^n.
$$
This is monic, since the subtracted <polynomial> has lower degree, and is orthogonal to every lower-degree <polynomial>. If $\phi_n,\psi_n$ were two such <monic polynomials>, their difference would lie in $V_{n-1}$ and be orthogonal to that space. Taking its <inner product> with itself gives zero, hence the difference vanishes. This proves existence and uniqueness for every $n$, with $\phi_0=1$.
Odd moments vanish because the weight is even; the even moments are
$$
\int_{\mathbb R}e^{-|x|}x^{2k}dx
=2\int_0^\infty e^{-x}x^{2k}dx=2(2k)!.
$$
Consequently $\phi_1=x$, while a monic quadratic orthogonal to $x$ has no linear term, and orthogonality to one gives its constant term $-4/2=-2$. Thus
$$
\boxed{\phi_0=1,\qquad\phi_1=x,\qquad\phi_2=x^2-2.}
$$
Finally $(-1)^n\phi_n(-x)$ is also monic and orthogonal to every lower-degree <polynomial>, by the change of variable $x\mapsto-x$. Uniqueness implies $\phi_n(-x)=(-1)^n\phi_n(x)$. In particular $\phi_{2007}$ is odd and
$$
\boxed{[x^{2000}]\phi_{2007}=0.}
$$
This is <parity of monic orthogonal polynomials for an even weight>.
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