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 $q(x)=a+bx+c{x}^2$ (that is, a general quadratic), then ${\int }_{-1}^{1}a+bx+c{x}^{2}dx=2a+\frac{2c}{3}$, and ${\Lambda }_{1}q(-\sqrt{\frac{3}{5}})+{\Lambda }_{2}q\left(0\right)+{\Lambda }_{3}q\left(\sqrt{\frac{3}{5}}\right)=\frac{5}{9}(a-b\sqrt{\frac{3}{5}}+\frac{3c}{5})+\frac{8a}{9}+\frac{5}{9}(a+b\sqrt{\frac{3}{5}}+\frac{3c}{5})=\frac{10a}{9}+\frac{8a}{9}+\frac{2c}{3}=2a+\frac{2c}{3}$ 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 $x^5$.
${\int }_{-1}^1{x}^{3}dx=0$ and $\frac{5}{9}(-\frac{3}{5}\sqrt{\frac{3}{5}})+\frac{5}{9}\left(\frac{3}{5}\sqrt{\frac{3}{5}}\right)=0$.
${\int }_{-1}^1{x}^{4}dx=\frac{2}{5}$ and $\frac{5}{9}\left(\frac{9}{25}\right)+\frac{5}{9}\left(\frac{9}{25}\right)=\frac{2}{5}$.
${\int }_{-1}^1{x}^{5}dx=0$ and $\frac{5}{9}(-\frac{9}{25}\sqrt{\frac{3}{5}})+\frac{5}{9}(-\frac{9}{25}\sqrt{\frac{3}{5}})=0$.
${\int }_{-1}^1{x}^{6}dx=\frac{2}{7}$ 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.