若关于$x$的不等式组$\begin{cases}x-a>2 \\ b-2x>0 \\ \end{cases}$的解集是$-1<{}x<{}1$，则${{\left( a+b \right)}^{2019}}=$ ．
解不等式组$\begin{cases}x-a>2 \\ b-2x>0 \\ \end{cases}$，得$a+2<{}x<{}\frac{b}{2}$，
$\because $不等式组$\begin{cases}x-a>2 \\ b-2x>0 \\ \end{cases}$的解集为$-1<{}x<{}1$，
$\therefore a+2=-1$，$\frac{b}{2}=1$
$\therefore a=-3$，$b=2$，
$\therefore {{\left( a+b \right)}^{2019}}={{\left( -3+2 \right)}^{2019}}={{\left( -1 \right)}^{2019}}=-1$.