= Solution
On $D_+(x_3)$, write $r=x_0/x_3$, $q=x_1/x_3$, and $s=x_2/x_3$. The equations include $q=\pi s^2$ and $r=s^3$; after making these substitutions, the remaining equations are identities. Therefore
$$
\boxed{D_+(x_3)\cong\operatorname{Spec}R[s]=\mathbb A_R^1.}
$$
The other chart is
$$
U=D_+(x_0)=\operatorname{Spec}B,\qquad B=R[a,b,c]/(a^2-\pi^2b,\ ac-\pi b^2,\ ab-\pi c,\ c^2-b^3).
$$
There is a homomorphism $B\to R[t]$ defined by $a\mapsto\pi t$, $b\mapsto t^2$, $c\mapsto t^3$. To prove injectivity rather than merely exhibit a parametrization, reduce monomials using the four relations. Powers of $a$ reduce to at most one $a$; any remaining product of $a$ with $b$ or $c$ eliminates $a$; powers of $c$ reduce to at most one $c$. Thus every element is an $R$-linear combination of
$$
1,\quad a,\quad b^n\ (n\ge1),\quad cb^n\ (n\ge0).
$$
Their images are $1,\pi t,t^{2n},t^{2n+3}$, with pairwise distinct degrees. Since $R$ is a domain and $\pi\ne0$, these images are $R$-linearly independent. Hence they are a <basis>, the map is injective, and
$$
\boxed{B\cong R[\pi t,t^2,t^3]\subseteq R[t].}
$$
In particular $B$ is torsion-free, indeed free, as an $R$-module.
The two charts cover $X$ by the prime-ideal argument in part i. Both coordinate <rings> are torsion-free over the <discrete valuation ring> $R$, hence flat: over a <principal ideal domain> a <torsion-free module> is a filtered union of finitely generated free submodules, and filtered colimits preserve flatness. Thus the morphism $X\to\operatorname{Spec}R$ is flat on an open cover and therefore flat everywhere. It is proper because $X$ is a closed subscheme of the proper <projective space> $\mathbb P_R^3$. Consequently \b[$X$ is both proper and flat over $R$].
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