John Lesieutre from State College Area High School, Pennsylvania, USA and Marcos Charalambides from Cyprus sent in excellent solutions to this problem.

Using $q(x)=1$ we get $2=\Lambda_1+\Lambda_2+\Lambda_3$.

Using $q(x)=x$ we get $0=\Lambda_1\sqrt{\frac{3}{5}}+\Lambda_3\sqrt{\frac{3}{5}}$ so $\Lambda_1=\Lambda_3$.

Using $q(x)=x^2$ we get $\frac{2}{3}=\frac{3\Lambda_1}{5}+\frac{3\Lambda_3}{5}= \frac{6\Lambda_1}{5}$ so $\Lambda_1=\frac{5}{9}=\Lambda_3$ and $\Lambda_2=\frac {8}{9}$.

Now if $q(x)=a+b x+c x^2$ (that is, a general quadratic), then $\int_{-1}^1 a+b x+c x^2 \mathrm{d}x=2a+\frac{2c}{3}$, and $\Lambda_1 q\left(-\sqrt{\frac{3}{5}}\right)+\Lambda_2 q\left(0\right)+\Lambda_3 q\left(\sqrt{\frac{3}{5}}\right)=\frac{5}{9}\left(a-b\sqrt{\frac{3}{5}}+\frac{3 c}{5}\right)+\frac{8a}{9}+\frac{5}{9}\left(a+b\sqrt{\frac{3}{5}}+\frac{3c}{5}\right) =\frac{10a}{9}+\frac{8a}{9}+\frac{2c}{3} =2a+\frac{2c}{3} $ so the formula works for all quadratics.

Using the same idea as above, to check that it works for cubics, quartics and quintics, we only have to check it for $x^3$, $x^4$ and $x^5$.

$\int_{-1}^1 x^3\mathrm{d} x = 0$ and $\frac{5}{9}\left(-\frac{3}{5}\sqrt{\frac {3}{5}}\right)+\frac{5}{9}\left(\frac{3}{5}\sqrt{\frac{3}{5}}\right)=0$.

$\int_{-1}^1 x^4\mathrm{d} x=\frac{2}{5}$ and $\frac{5}{9}\left(\frac{9}{25}\right) +\frac{5}{9}\left(\frac{9}{25}\right)=\frac{2}{5}$.

$\int_{-1}^1 x^5\mathrm{d} x=0$ and $\frac{5}{9}\left(-\frac{9}{25}\sqrt{\frac{ 3}{5}}\right)+\frac{5}{9}\left(-\frac{9}{25}\sqrt{\frac{3}{5}}\right)=0$.

$\int_{-1}^1 x^6\mathrm{d} x=\frac{2}{7}$ but $\frac{5}{9}\left(\frac{27}{125} \right)+\frac{5}{9}\left(\frac{27}{125}\right)=\frac{6}{25}$.

So the formula works for cubics, quartics and quintics, but not for higher powers.