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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.

 

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