= Solution
Here the inverse in the <elliptic curve group law> is $(x,y)\mapsto(x,-y-1)$. For $P_0+P_1$, the line is $y=2-2x$. Substitution in the <Weierstrass equation of an elliptic curve> gives
$$
x^3-7x+6-\bigl((2-2x)^2+(2-2x)\bigr)=x(x-1)(x-3).
$$
The third intersection is $(3,-4)$, so reflection gives
$$
\boxed{P_0+P_1=(3,3).}
$$
For the difference, first take $-P_1=(1,-1)$. The line through $P_0$ and $-P_1$ is $y=2-3x$, and now the intersection polynomial is $x(x-1)(x-8)$. Its third point is $(8,-22)$, giving
$$
\boxed{P_0-P_1=(8,21).}
$$
Finally, implicit <differentiation> gives $(2y+1)y'=3x^2-7$. The <tangent line> at $P_1$ therefore has slope $-4$ and equation $y=4-4x$. Its intersection polynomial factors as $(x-1)^2(x-14)$. The double root records the <intersection multiplicity> of the tangent, and the remaining intersection is $(14,-52)$. Hence
$$
\boxed{2P_1=(14,51).}
$$
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