= Solution
For <two-node Gaussian quadrature with linear weight>, the monic node <polynomial> $q(x)=x^2-sx+p$ must be orthogonal to $1$ and $x$ for weight $x$. Indeed, exactness through degree three makes the quadrature of $q$ and $xq$ vanish at its nodes. The moment equations are
$$
\frac14-\frac s3+\frac p2=0,\qquad
\frac15-\frac s4+\frac p3=0,
$$
so $s=6/5$ and $p=3/10$. Thus the <Gaussian quadrature> nodes, labelled in increasing order, are
$$
\boxed{x_1=\frac{6-\sqrt6}{10},\qquad x_2=\frac{6+\sqrt6}{10}.}
$$
Integrating the linear <Lagrange interpolation> cardinal functions gives the weights
$$
\boxed{a_1=\frac{x_2/2-1/3}{x_2-x_1},\qquad
a_2=\frac{1/3-x_1/2}{x_2-x_1}.}
$$
They are positive, sum to $1/2$, and reproduce the first moment $1/3$. Any cubic <polynomial> has remainder of degree at most one after division by $q$; orthogonality annihilates the other part both in the integral and at the nodes. This proves degree-three exactness, not merely the two moment conditions for the weights.
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