$\because x^{2}-2\left(k-1\right)x+2k^{2}-12k+17=0$，两根为$x_{1}$，$x_{2}$，
$\therefore \Delta =\left[-2\left(k-1\right)\right]^{2}-4\times 1\times (2k^{2}-12k+17)=-4k^{2}+40k-64\geqslant 0$
$\therefore k^{2}-10k+16\leqslant 0$，
由图象可知，$k^{2}-10k+16\leqslant 0$的解集为$2\leqslant k\leqslant 8$，

$\because {x}_{1}^{2}+{x}_{2}^{2}=(x_{1}+x_{2})^{2}-2x_{1}x_{2}=16k-30 \gt 0$.
$\therefore {x}_{1}^{2}+{x}_{2}^{2}$的值随着$k$的增大而增大，
$\therefore$当$k=2$时，${x}_{1}^{2}+{x}_{2}^{2}$取最小值为$16\times 2-30=2$，
当$k=8$时，${x}_{1}^{2}+{x}_{2}^{2}$取最大值为$16\times 8-30=98$，
$\therefore {x}_{1}^{2}+{x}_{2}^{2}$的最大值为$98$，最小值为$2$.

$\because x^{2}-2\left(k-1\right)x+2k^{2}-12k+17=0$，两根为$x_{1}$，$x_{2}$，
$\therefore \Delta =\left[-2\left(k-1\right)\right]^{2}-4\times 1\times (2k^{2}-12k+17)=-4k^{2}+40k-64\geqslant 0$
$\therefore k^{2}-10k+16\leqslant 0$，
由图象可知，$k^{2}-10k+16\leqslant 0$的解集为$2\leqslant k\leqslant 8$，

$\because {x}_{1}^{2}+{x}_{2}^{2}=(x_{1}+x_{2})^{2}-2x_{1}x_{2}=16k-30 \gt 0$.
$\therefore {x}_{1}^{2}+{x}_{2}^{2}$的值随着$k$的增大而增大，
$\therefore$当$k=2$时，${x}_{1}^{2}+{x}_{2}^{2}$取最小值为$16\times 2-30=2$，
当$k=8$时，${x}_{1}^{2}+{x}_{2}^{2}$取最大值为$16\times 8-30=98$，
$\therefore {x}_{1}^{2}+{x}_{2}^{2}$的最大值为$98$，最小值为$2$.