a) unless I am mistaken, since the space does not have a unity it cannot be a subspace. edit: nm I think that a submodule need not have a unity so I think it is a subspace now

b) not closed under pointwise addition at x = 0 so not a vector space

c) Let V = {even functions}. ill get back to you on this one. edit: the guy below me gave a good answer for this one

d) as a subset of polynomial space, the set of V ={polynomials of degree 2} does NOT determine a vector space because there is no 0 vector in this space. the 0 vector ( f(x) = 0) is a constant function which has degree 0 and hence is not a member of the set of polynomials of deg 2. Also a subspace is a vector space in its own right, so it has to have a unity, i.e. an element 1 s.t. 1f = f for all f in the subspace. this would have to be f(x) = 1, which doesn't exist in V either.