$L_1$ and $L_2$ are two arbitrary languages
Option 1. $\text{if}\; L_1.L_2\;\text{ is regular then }L_2.L_1\;\text{ is also regular}$
Let $L_1=\{a^p \; \mid \; \text{p is prime}\}$, and $L_2=\{a^nb^m \; \mid \; m,n\geq 0\}$
then $L_1.L_2=\{a^ib^j \; \mid \; i\geq 2,j\geq 0\}$ is Regular.
and $L_2.L_1=\{a^ib^ja^p \; \mid \; i,j\geq 0,\text{p is prime}\}$ is Non-regular.
Option 2. $L_1 = L_2 \;\text{iff}\; L_1 \backslash L_2 = \phi \; \text{and}\; L_2 \backslash L_1 = \phi$
$L_1 \backslash L_2 = \phi$ means all strings of $L_1$ are present in $L_2$ , and
$L_2 \backslash L_1 = \phi$ means all strings of $L_2$ are present in $L_1$
option3. $\text{if}\; L\subseteq \Sigma^* \;\text{and}\; L\; \text{is finite, then}\; \Sigma^* \backslash L \;\text{is regular}$
$L$ is finite, means $L$ is regular , so $\Sigma^* \backslash L = \Sigma^* - L $ is also regular
Only i) is incorrect
