I'm going to take a guess and assume that $\equiv_L$ is defined as the set of all pairs $(x,y) \in \Sigma^* \times \Sigma^*$ such that either (i) $x \in L$ and $y \in L$, or (ii) $x \not\in L$ and $y \not\in L$.
In this case $\Sigma^* / \equiv_L$ contains two equivalence classes, namely $L$ itself and its complement $\Sigma^* \setminus L$.
A regular expression for $L$ is $1^*(01^*01^*)^*1$. Essentially we make sure that $0$s always come in pairs and that the last character is a $1$.
A regular expression for $\Sigma^* \setminus L$ is $\varepsilon \mid (0\mid1)^*0 \mid 1^*(01^*01^*)^*01^*$.
The first part of the regular expression (i.e., $\varepsilon$) matches the empty word. The second part (i.e., $(0\mid1)^*0$) matches all words that end with $0$, and the third part (i.e., $1^*(01^*01^*)^*01^*$) matches all words with an odd number of $0$s.
As Nir Shahar points out, another definition of $\equiv_L$ could be $x \equiv_L y$ if and only if $\not\exists z \in \mathbb \Sigma^*$ such that either (i) $xz \in L$ and $yz \not\in L$ or (ii) $xz \not\in L$ and $yz \in L$. This is the definition used in the Myhill-Nerode theorem, and word $z$ satisfying either (i) or (ii) is called a distinguishing extension for $x$ and $y$.
In this case $\Sigma^* / \equiv_L$ contains the following classes:
- $C_1$: The set of all words that have an even number of $0$s and end with $1$, i.e., $L$ itself. A regular expression for $C_1$ is given above.
- $C_2$: The set of all words that have an odd number of $0$s. A regular expression for $C_2$ is given above.
- $C_3$: The set of all words that have an even number of $0$s but do not end with $1$. A regular expression for $C_3$ is $(1^* 0 1^* 0)^*$.
It is easy to see that $C_1, C_2, C_3$ partition $\Sigma^*$. We just need to show that (A) any two words in $x,y \in C_i$ satisfy $x \equiv_L y$, and (B) for $i \neq j$ there are two words $x \in C_i$, $y \in C_j$ that do not satisfy $x \equiv_L y$.
Regarding (A):
If $z$ were a distinguishing extensions for either two words in $x,y \in C_1$ or two words $x,y \in C_3$, then since at least one of $xz$ and $xz$ must belong to $L$ and we must have $z \neq \varepsilon$, $z$ must end with $1$ and must contain an even number of zeroes. However, any such $z$ is such that both $xz \in L$ and $xz \in L$.
If $z$ were a distinguishing extensions for two words in $x,y, \in C_2$, then since at least one of $xz$ and $xz$ must belong to $L$, $z$ must end with $1$ and must contain an odd number of zeroes. However, any such $z$ is such that both $xz \in L$ and $xz \in L$.
Regarding (B): $\varepsilon$ is a distinguishing extension for $1 \in C_1$ and $0 \in C_2$. $1$ is a distinguishing extension for $0 \in C_2$ and $\varepsilon \in C_3$. Finally, $\varepsilon$ is a distinguishing extension for $1 \in C_1$ and $\varepsilon \in C_3$.