Let L1 = {s ∈ (0 + 1)* | d(s) mod5 = 2}, we can construct the DFA for this which will have 5 states (remainders 0,1,2,3,4)
L2 = {s ∈ (0 + 1)* | d(s) mod7 = 4}, we can construct the DFA for this which will have 7 states (remainders 0,1,2,3,4,5,6)
Since L1 and L2 have DFAs, hence they are regular. So the resulting Language.
L = L1 ∩ L2 (compliment) must be regular (by closure properties, INTERSECTION of two regular languages is a regular language).