Explanation : I {ambncpdq | m + p = n + q} is clearly CFL since, we can rearrange the equation as m - n + p - q = 0 which can be done by push, pop, and pop and check if stack is empty at
end.
II {ambncpdq | m = n and p = q} is clearly CFL since, one comparison at a time can be done by pda
III {ambncpdq | m = n = p and p ≠ q} is not CFL since m = n = p is a double comparison which cannot be done by PDA.
IV {ambncpdq | mn = p + q} is not a CFL, since mn involves multiplying number of a’s and number b’s which cannot be done by a PDA.
So, only I and II are CFL’s.