= Solution
The discriminant is $-64D^3$, which is nonzero modulo three because $D\equiv2\pmod3$. Thus there is <good reduction> at three. For every $x\in\mathbb F_3$, the reduced right side is $x^3-x=0$, so its only affine points are $(0,0),(1,0),(2,0)$. Together with the identity they form $C_2\times C_2$.
By part (ii), the rational <torsion subgroup> injects into this four-element <group> and hence has exponent two. Over $\mathbb Q$, the two-torsion points have $Y=0$, requiring $X(X^2+D)=0$. Since $D>0$, the only rational root is zero. Therefore
$$
\boxed{E(\mathbb Q)_{\mathrm{tors}}=\{O,(0,0)\}\cong C_2.}
$$
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