= Solution
Use $P=(0,0)$ in the family from part (b). For order four, take $a=b=1$, giving
$$
\boxed{E_4:y^2+xy+y=x^3+x^2,\qquad\operatorname{ord}(P)=4.}
$$
Its discriminant is $-15$, and $2P=(-1,0)$ is a nonzero point equal to its own inverse. Thus $4P=O$ and $2P\ne O$.
For order five, take $a=2,b=1$:
$$
\boxed{E_5^{\mathrm{example}}:y^2+2xy+y=x^3+x^2,\qquad\operatorname{ord}(P)=5.}
$$
The discriminant is $-11$, and $a=b+1$ gives exact order five by part (b).
For order six, take $a=0,b=-2$:
$$
\boxed{E_6:y^2-2y=x^3-2x^2,\qquad\operatorname{ord}(P)=6.}
$$
Its discriminant is $80$. Here $2P=(2,2)$ and $3P=(1,1)$; negation is $(x,y)\mapsto(x,2-y)$, so $3P$ is nonzero 2-torsion. Neither $2P$ nor $3P$ is $O$, excluding orders two and three, while $6P=O$. All coefficients and points are rational, and the nonzero discriminants certify genuine <elliptic curves>.
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