= Solution
With $x=\cos\theta$, the weighted <inner product> for the <Chebyshev polynomials> becomes
$$
\int_{-1}^1\frac{T_n(x)T_m(x)}{\sqrt{1-x^2}}\,dx=\int_0^\pi\cos(n\theta)\cos(m\theta)\,d\theta.
$$
The identity $2\cos(n\theta)\cos(m\theta)=\cos((n-m)\theta)+\cos((n+m)\theta)$ and the vanishing of $\int_0^\pi\cos(k\theta)\,d\theta$ for positive integers $k$ give zero when $n\neq m$, $\pi/2$ when $n=m>0$, and $\pi$ when $n=m=0$. Thus
$$
\boxed{\int_{-1}^1\frac{T_nT_m}{\sqrt{1-x^2}}\,dx=\frac\pi2\delta_{nm}(1+\delta_{n0}).}
$$
For <Gaussian quadrature>, choose the $n$ distinct nodes as the zeros of the degree-$n$ <orthogonal polynomial> for the weight, here $T_n$, and choose the weights as the weighted integrals of their <Lagrange interpolation> basis functions. If a <polynomial> $p$ has degree at most $2n-1$, divide it as $p=qT_n+r$, where both $q$ and $r$ have degree at most $n-1$. Orthogonality makes the integral of $qT_n$ zero, while its quadrature sum also vanishes; interpolation integrates $r$ exactly. This proves degree $2n-1$ exactness. It is optimal because the squared nodal <polynomial>, of degree $2n$, has zero quadrature sum and strictly positive integral.
For $n=3$, $T_3(x)=4x^3-3x$, so the nodes are $-\sqrt3/2,0,\sqrt3/2$. Their interpolation basis functions are
$$
\ell_-(x)=\frac23x\left(x-\frac{\sqrt3}{2}\right),\quad \ell_0(x)=1-\frac43x^2,\quad \ell_+(x)=\frac23x\left(x+\frac{\sqrt3}{2}\right).
$$
The weighted zeroth and second moments are $\pi$ and $\pi/2$, and the first moment is zero. Integrating these three basis functions gives equal weights $\pi/3$, hence the <Chebyshev–Gauss quadrature> rule
$$
\boxed{\int_{-1}^1\frac{f(x)}{\sqrt{1-x^2}}\,dx\approx\frac\pi3\left[f\!\left(-\frac{\sqrt3}{2}\right)+f(0)+f\!\left(\frac{\sqrt3}{2}\right)\right],}
$$
which is exact for every <polynomial> of degree at most $5$.
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