$(01+1)^*$ represents the language consisting precisely of those words in which every $0$ (if there is one) is immediately followed by a $1$. This excludes all words containing $00$ and all words that end in $0$, and nothing else. An easy way to see this is to realize that $(01+1)^*$ describes the language the same language that you get if you start with $\{1,2\}^*$ (corresponding to the regular expression $(2+1)^*$) and replace each $2$ in every word by $01$. The resulting word clearly must end in $1$ and clearly cannot contain $00$, but itβs not hard to show that it contains everything else.
I assume that your other regular expression is supposed to be $(0+1)^*[0+00(0+1)^*]$; the two $(0+1)^*$ expressions allow the generation of arbitrary strings, so the language is everything of the form $u0$ with $u\in\{0,1\}^*$ together with everything of the form $u00v$ with $u,v\in\{0,1\}^*$ $-$ in other words, everything that either ends in $0$ or contains $00$, precisely the complement of the first language.