解：??$(1)b＜c＜a，$??理由如下：
因為??$a={2}^{-44444}={({2}^{-4})}^{11111}={(\frac 1 {16})}^{11111}，$????$b={3}^{-33333}={({3}^{-3})}^{11111}={(\frac 1 {27})}^{11111}，$??
??$c={5}^{-22222}={({5}^{-2})}^{11111}={(\frac 1{25})}^{11111}，$??
又因為??$\frac 1 {27}＜\frac 1 {25}＜\frac 1 {16}，$??
所以??${(\frac 1 {27})}^{11111}＜{(\frac 1 {25})}^{11111}＜{(\frac 1 {16})}^{11111}，$??即??$b＜c＜a$??
??$(2)①$??當??$2x+3=1，$??則??$x=-1；$??
②當??$2x+3=-1$??且??$x+2020$??為偶數(shù)，則??$x=-2；$??
③當??$x+2020=0，$??則??$x=-2020$??
綜上所述，當??$x=-1$??或??$-2$??或??$-2020$??時，??${(2x+3)}^{x+2020}=1$??成立
