$$\begin{gathered} \left\{\begin{array}{l}2x+y=19\\ 3x-2y=11\end{array}\right.\iff \left\{\begin{array}{l}4x+2y=38\\ 3x-2y=11\end{array}\iff \left\{\begin{array}{l}7x=49\\ 3x-2y=11\end{array}\iff \left\{\begin{array}{l}x=7\\ 21-2y=11\end{array}\right.\right.\right. \\ \iff \left\{\begin{array}{l}x=7\\ y=5\end{array}\right. \end{gathered}$$

Vậy phương trình có nghiệm (x;y)=(7;5)

Theo Vi-ét:

$\left\{\begin{array}{l}{x}_1+x_2=\frac{-b}{a}=4\\ x_1{x}_2= \frac{c}{a}=3m-1\end{array}\right.$

Ta có: 

$$\begin{gathered} {x_1}^2+{x_2}^2 =15 \\ \iff {x_1}^2+2x_1{x}_2+{x_2}^2-2x_1{x}_2=15 \\ \iff {(x_1+x_2)}^2-2x_1{x}_2=15 \\ \Rightarrow 4^2-2(3m-1)=15 \\ \iff 16-6m+2=15 \\ \iff -6m=-3 \\ \iff m=\frac{1}{2} \end{gathered}$$

Vậy $m=\frac{1}{2}$ thì phương trình thỏa mãn ${x_1}^2+{x_2}^2=15$