regex - difference between $\Lambda+(a+b)^*b$ and $(b+aa^*b)^*$
Yes, the regular expressions $r_1=\Lambda+(a+b)^*b$ and $r_2=(b+aa^*b)^*$ are equivalent.  There are several ways of doing this.

Just observe that both of them match the empty string and all strings ending with $b$.

Convert them to automata and then produce automata for $L(r_1)\setminus L(r_2)$ and $L(r_2)\setminus L(r_1)$. Minimize both automata and find that they both accept the empty language.  (Actually, you'll find that you make a ton of mistakes and at least one of the automata accepts something, and at least one of the minimizations is wrong. But, if you were a computer, this wouldn't be an issue.)

Clearly both regular expressions match the empty string, so let's show that they match the same non-empty strings, too.  Suppose that $w\in L(r_2)\setminus\{\Lambda\}$. Then we can write $w=w_1\dots w_\ell$ where each $w_i$ is a string that either matches $b$ or $aa^*b$. In particular, $w_\ell$ must match one of these things, so the last character of $w_\ell$ is a $b$, so $w\in L(r_1)$.
Conversely, suppose that $w\in L(r_1)\setminus\{\Lambda\}$. Write $w=s_1\dots s_n$, where each $s_i\in\{a,b\}$. Suppose there are $\ell$ $b$s in $w$ and let these be at positions $p_1, \dots, p_\ell$. Because $w\in L(r_1)$, $p_\ell=n$ (the last character is $b$).  For convenience, let $p_0=1$. We can now write $w=w_1\dots w_\ell$, where $w_i=s_{p_{i-1}+1}\dots s_{p_i}$. For each $i\in\{1, \dots, \ell\}$, we have $w_i=a^{p_i-p_{p-1}-1}b$, so each $w_i$ matches either the regular expression $b$ or $aa^*b$. Therefore, $w$ matches $(b+aa^*b)^*= r_1$.


By the way, a fourth equivalent regular expression is $(a^*b)^*$, which you can see by observing that the inner part of $r_1$ is equivalent to $b+a^+b$, which is $a^*b$.