a) { b1000+n | n>=0 } ∪ {ab1000+n | n>=0 } ∪ {a2b1000+n | n>=0 } ∪ .............{a1000b1000+n | n>=0 }
Reg ∪ Reg ∪ Reg .....∪ Reg = Reg // iff it's not infinite union
b) Comparison between a and b , certainly CFL // Remember na(w)=nb(w) or na(w)-nb(w)=0 well known CFL
c) It's (a+b)* so Reg
So in that way L1 and L3 are Regular languages.