= Solution
By the <spectral theorem for real symmetric matrices>, write $A=O^tDO$ with $O$ <orthogonal matrix>[orthogonal] and $D=\operatorname{diag}(\xi_1,\ldots,\xi_n)$. Part a and the tensor-product property of the <Fourier transform> give
$$
\mathcal F\left[e^{iDx\cdot x/2}\right](\lambda)
=\prod_{j=1}^n
\left(
\sqrt{\frac{2\pi}{|\xi_j|}}
e^{i\pi\operatorname{sgn}(\xi_j)/4}
e^{-i\lambda_j^2/(2\xi_j)}
\right).
$$
Thus
$$
\mathcal F\left[e^{iDx\cdot x/2}\right](\lambda)
=\sqrt{\frac{(2\pi)^n}{|\det D|}}
\exp\left[
\frac{i\pi}{4}\operatorname{sgn}(D)
-\frac i2D^{-1}\lambda\cdot\lambda
\right].
$$
Applying the pullback rule from part c to the orthogonal change of variables, for which $|\det O|=1$, replaces $D$ by $A$ and $D^{-1}$ by $A^{-1}$. Since determinant and <signature> are invariant under orthogonal conjugation,
$$
\boxed{
\left[e^{iAx\cdot x/2}\right]^{\widehat{}}(\lambda)
=\sqrt{\frac{(2\pi)^n}{|\det A|}}
\exp\left[
\frac{i\pi}{4}\operatorname{sgn}(A)
-\frac i2(A^{-1}\lambda)\cdot\lambda
\right]}.
$$
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