$L$ is a context free language over $\{a,b\}$
$L_1 = L - \{xyx \mid x,y \in \{a,b\}^* \}$
$= L - \{ \text{all strings over} \{a,b\} \}$ [ Note: all strings can be generated from y by putting $x= \epsilon$]
$= L - (a+b)^* = \{\}$ , empty set. [Note : $L_1 - L_2 = \{ \text{string in $L_1$ but not in L2 }\}$ ]
So , $L_1$ is a Regular Language.
$L$ is a context free language over $\{a,b\}$
$L_2 = L \cdot L$
Context free languages are closed under Concatenation.
So, $L_2$ is Context Free Language.
Option C is correct.