= Solution
For this <Weierstrass equation of an elliptic curve>, the inverse is $(x,y)\mapsto(x,13-y)$. At a point with $2y-13\ne0$, implicit differentiation gives the tangent slope
$$
m=\frac{3x^2+2x+13}{2y-13}.
$$
If the tangent line is $y=mx+b$, substitution into the equation shows that the three intersection abscissae sum to $m^2-1$. Two of them equal the tangency abscissa $x$, so the third is $x_3=m^2-1-2x$. Reflecting the third intersection by the inverse formula gives the <elliptic curve group law from Riemann-Roch> doubling formula
$$
x(2R)=m^2-1-2x(R),\qquad
y(2R)=13-y(R)+m\bigl(x(R)-x(2R)\bigr).
$$
For $P$, the slope is $13/(-13)=-1$, giving $x(2P)=0$ and $y(2P)=13$. For $Q$, the slope is $(12-4+13)/(6-13)=-3$, giving $x(2Q)=9-1+4=12$ and $y(2Q)=13-3+(-3)(-2-12)=52$. Hence
$$
\boxed{2P=(0,13)=-P,\qquad 2Q=(12,52).}
$$
In particular $P$ has order exactly three: it is nonidentity and $2P=-P$.
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