The key here is to know that taking a derivative is a linear operation. For the set in question to be a subspace it needs to be closed under linear operations and must contain the 0 vector (in this case, the constant function f=0). Since taking a derivative is a linear operation, I, II, and III are all obviously closed under linear operations. (They are also all linear ODEs if that helps to convince you). The catch with III is that it doesn't contain the 0 function, so it cannot be a subspace.
Edit: One other thing I thought of. If you've done some ODEs before you'll know that one way to solve equations like these is to consider the derivatives instead as applications of a differential operator. So equation I, for example, would be D^2 - 2D + 3 = 0 if D is our operator. Then we can solve for D just like in a polynomial. If the polynomial has real roots say, A and B, then the solution to the differential equation is C(e^At) + D(e^Bt) where C and D are arbitrary constants. But if the polynomial has imaginary roots (likely the imaginary number you calculated) then we need to use Euler's identity e^it = cost + isint to make sense of it. So now the solutions will take the same form as before (C(e^At) + D(e^Bt)) but if A and B have some imaginary part then we need to break the solution down using Euler's identity and take the real part of the whole expression which will likely be some sum of sines and cosines of different periods. Note that since these are linear differential equations, taking just the real part of the solution won't stop it from satisfying the differential equation. So in trying to calculate a solution you probably weren't far off, you just needed to use Euler's identity and take the real part.