解:$ (1) $把方程$②$变形为$3(3x - 2y)+2y = 19 ③,$
把①代入③,得$15 + 2y = 19,$
所以$y = 2,$
把$y = 2$代入$①,$得$x = 3,$
则方程组的解为$\begin {cases}x = 3 \\y = 2\end {cases}。$
$ (2) $由$3x^2-2xy + 12y^2=47,$得
$3(x^2+4y^2) = 47 + 2xy,$
即$x^2+4y^2=\frac {47 + 2xy}{3},$
由$2x^2+xy + 8y^2=36,$得
$2x^2+8y^2=36 - xy,$
则$2×\frac {47 + 2xy}{3}=36 - xy,$解得$xy = 2,$
则$x^2+4y^2=17。$