because L1 can be rewritten as,

L1 = {w | w $\epsilon$ (a+b)* , |w|mod3 = 0 }, which is regular.

language will be ((a+b)*)^{n}.((a+b)*)^{n}.((a+b)*)^{n}

Cancatenation of 3 equal length string

yes string will be mod3 and regular

$L_1$ can be written simply as $\{ x : x \in (a+b)^*, |x|mod3 = 0\}$.

Therefore it is regular.