= Solution
Write the equation as $y^2=x^3-5x^2+4x$. Its <elliptic curve group law>[duplication formula] uses slope $m=(3x^2-10x+4)/(2y)$ and $x(2P)=m^2+5-2x$. At the three listed points the slopes are respectively $i$, $-1+i\sqrt3$, and $2+\sqrt3$. Substitution gives
$$
\boxed{2P_1=(0,0),\qquad2P_2=(1,0),\qquad2P_3=(4,0).}
$$
These are nonidentity points of order two, so each listed point has exactly order four.
For $p\ne2,3$ the three distinct <roots of a polynomial>[roots] zero, one and four give the full two-<torsion subgroup> of order four over $\mathbb F_p$. At least one of $-1,3,-3$ is a square: if neither of the first two is, their product is. If $-1$ is a square, the first listed point exists over the <field>. If three is a square, the third exists. If $-3$ is a square, the second can be written $(1-\sqrt{-3},3+\sqrt{-3})$ and exists without requiring the two separate square <roots of a polynomial>[roots]. The duplication computations remain valid after reduction; their ordinates do not vanish unless $p=2$ or three. Thus a rational <2-torsion>[two-torsion] point has a half over the <field>. Together with a different <2-torsion>[two-torsion] point this generates a <subgroup> of order eight. Consequently $\boxed{8\mid\#E(\mathbb F_p)}$ for every stated <prime number>[prime].
Over the rationals there are already four <2-torsion>[two-torsion] points, including the identity. Counting at five gives eight points: the abscissas zero, one and four give one ordinate each, two and three give two each, and there is the identity. Thus <torsion subgroup>[rational torsion] has order dividing eight.
There are no <rational points> of order four. For a curve $y^2=\prod_j(x-e_j)$, translating $e_i$ to zero and using the <elliptic curve group law>[duplication formula] shows that a half of $(e_i,0)$ must satisfy $(x-e_i)^2=(e_i-e_j)(e_i-e_k)$. For the three <roots of a polynomial>[roots] here, the candidate abscissas are $\pm2$, $1\pm\sqrt{-3}$, and $4\pm2\sqrt3$. The last two pairs are nonrational; at $x=2,-2$ the ordinate squares are $-4,-36$, also impossible over $\mathbb Q$. A group of order eight containing this two-<torsion subgroup> would contain an element of order four, because an <elliptic curve> has only four points killed by two. Hence
$$
\boxed{E(\mathbb Q)_{\mathrm{tors}}\cong C_2\times C_2,\qquad\#E(\mathbb Q)_{\mathrm{tors}}=4.}
$$
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