$\left(1\right)2\left(x-y\right)^{2}-5\left(x-y\right)^{2}+\left(x-y\right)^{2}$

$=\left(2-5+1\right)\left(x-y\right)^{2}$

$=-2\left(x-y\right)^{2}$，

故答案为：$-2\left(x-y\right)^{2}$；

$(2)$①$\because a^{2}+a=1$，

$\therefore 2a^{2}+2a+2020$

$=2(a^{2}+a)+2020$

$=2\times 1+2020$

$=2+2020$

$=2022$，

故答案为：$2022$；

②$\because a+b=-3$，

$\therefore 5\left(a+b\right)+7a+7b+11$

$=5\left(a+b\right)+7\left(a+b\right)+11$

$=12\left(a+b\right)+11$

$=12\times \left(-3\right)+11$

$=-36+11$

$=-25$，

故答案为：$-25$；

$(3)\because a^{2}-2ab=-5$，

$\therefore 2a^{2}-4ab=-10$，

$\because ab+2b^{2}=-3$，

$\therefore 2a^{2}-4ab+ab+2b^{2}=-10+\left(-3\right)=-13$，

$\therefore 2a^{2}-3ab+2b^{2}=-13$.

$\left(1\right)2\left(x-y\right)^{2}-5\left(x-y\right)^{2}+\left(x-y\right)^{2}$

$=\left(2-5+1\right)\left(x-y\right)^{2}$

$=-2\left(x-y\right)^{2}$，

故答案为：$-2\left(x-y\right)^{2}$；

$(2)$①$\because a^{2}+a=1$，

$\therefore 2a^{2}+2a+2020$

$=2(a^{2}+a)+2020$

$=2\times 1+2020$

$=2+2020$

$=2022$，

故答案为：$2022$；

②$\because a+b=-3$，

$\therefore 5\left(a+b\right)+7a+7b+11$

$=5\left(a+b\right)+7\left(a+b\right)+11$

$=12\left(a+b\right)+11$

$=12\times \left(-3\right)+11$

$=-36+11$

$=-25$，

故答案为：$-25$；

$(3)\because a^{2}-2ab=-5$，

$\therefore 2a^{2}-4ab=-10$，

$\because ab+2b^{2}=-3$，

$\therefore 2a^{2}-4ab+ab+2b^{2}=-10+\left(-3\right)=-13$，

$\therefore 2a^{2}-3ab+2b^{2}=-13$.