= Solution
For
$$
E:y^2+y=x^3-4x+2,
$$
the negative of $(x,y)$ is $(x,-y-1)$. At $P=(0,1)$ the tangent slope is
$$
\lambda=\frac{3x(P)^2-4}{2y(P)+1}=-\frac43.
$$
The tangent is $y=1-4x/3$, so the addition formulas give
$$
x(2P)=\lambda^2-2x(P)=\frac{16}{9},
\qquad
y(2P)=-\lambda x(2P)-1-1=\frac{10}{27}.
$$
Thus
$$
\boxed{2P=\left(\frac{16}{9},\frac{10}{27}\right).}
$$
The line through $P=(0,1)$ and $Q=(-2,1)$ is $y=1$. Its third intersection has $x=2$, and reflection under $y\mapsto-y-1$ gives
$$
\boxed{P+Q=(2,-2).}
$$
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