This could be solved using some basic Regular Expression Identities.
1)$\phi$*=$\varepsilon$
2)R.$\phi$=$\phi$.R=$\phi$
3)R.$\varepsilon$=$\varepsilon$.R=R
Where R is some Regular Expression.
Let's find L3'=Complement of L3=a*-L3
Since LE is defined over Alphabet With only symbol a.
L3'=a*-{a,$\varepsilon$}
L3'={$\varepsilon$,$a$,${a^2}$,$a^3$,$a^4$,.......} - {$a$,$\varepsilon$}
L3'={$a^2$,$a^3$ ,$a^4$,$a^5$,.......}
L3'={$a^n$/ $n$>=2}
Now,Solving the Expression
L3'.L2.L1*+L1.L3'=L3'.$\varepsilon$.$\phi$*+$\phi$.L3'
=L3'.$\varepsilon$.$\varepsilon$ +$\phi$(using $1$ and $2$)
=L3'+$\phi$
=L3'
={$ a^2$/$n$>=2}
Thus ,option D holds.