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L1 is CSL.

L2 is CFL.

L3 is REG.

L4 = CSL.REG = REG ( FOR THE GIVEN LANG when we concatenate CSL with REG it makes $a^*b^*c^*$ which is REG;

Note: This is not a generalized case )

L5 = ( CSL $\cap$ CFL ) $\cup$ REG = CSL $\cup$ REG = REG ( reasoning same as above; whatever comes out of $\cap$ it will be REG anyway because of union with REG L)
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