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)