To answer Q1: There are trig identities at play. First, 0 is usually accepted under the definition of "homogeneous polynomial" (i.e., it's a polynomial whose coefficients are all zero) so there is no contradiction there. But for the case of the second moment being constant, we have the identity $\sin^2(\theta) + \cos^2(\theta) = 1$, so actually a constant is representable by a homogeneous trig polynomial of degree 2 polynomial, so again, no contradiction.

To answer Q1: There are trig identities at play. First, 0 is usually accepted under the definition of "homogeneous polynomial" (i.e., it's a polynomial whose coefficients are all zero) so there is no contradiction there. But for the case of the second moment being constant, we have the identity $\sin^2(\theta) + \cos^2(\theta) = 1$, so actually a constant is representable by a homogeneous trig polynomial of degree 2 polynomial, so again, no contradiction.

Edit: for Q2, The moment conditions can indeed be violated when the function is not Schwartz. A good reference for this is the paper:

Solmon, D. C. (1987). Asymptotic formulas for the dual Radon transform and applications. Mathematische Zeitschrift, 195(3), 321-343.

The main theorem of this paper shows that the inverse Radon transform maps any even Schwartz function $\phi$ over $\mathbb{S}^{d-1}\times \mathbb{R}$ to a $C^\infty$-smooth function on $\mathbb{R}^d$ that decays like $O(\|x\|^{-d})$ as $\|x\|\rightarrow \infty$, i.e., absolutely integrable along hyperplanes, but not necessarily absolutely integrable over all of $\mathbb{R}^d$. Here $\phi$ does not have to satisfy any of the moment conditions, even though the defining integrals are convergent since $\phi$ is Schwartz.