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