I will give a detailed hint for the first one, and address your concerns for the rest.

is the set of differentiable functions with continuous derivatives. The subset is the set of all such functions with non-negative derivative at 0. To see if is closed under vector addition, check that . This is the same as asking whether . Similarly for the scalar multiplication case, check if for any scalar c. HINT: what if is negative and ?

You didn't check if S is closed under vector addition.B. Being a subset of V, set S contains all real-defined functions defined on [a,b]. Also if f(a)=0 and f(b)=0, then S contains the neutral element of V. If we have some f(a)=f(b), multiplying both sides of the equation by the same scalar preserves the equality. (1),(2),(3) is satisfied so it is a subspace of V.

OK, although you used the variable twice for different things when calculating (2).C. p(0)=0 is the neutral element of V, and for polynomials in S, p(0)=0. If s(x)=ax^3+bx and t(x)=cx^3+dx, then (s+t)(x)=(a+c)x^3+(b+d)x which has the form p(x)=ax^3+bx so (2) is satisfied. If s(x)=ax^3+bx and k is a scalar, k(s(x)) = (ka)x^3 + (kb)x, which retains the form so (3) is satisfied. (1),(2),(3) is satisfied so it is a subspace of V.

yes.D. The 0 square matrix is symmetric and in V. For symmetric square matrices A and B, taking A+B by taking the sum of corresponding entries will give a symmetric matrix. Likewise if a symmetric matrix is multiplied by a scalar, its entries will still be symmetric. (1),(2),(3) is satisfied so it is a subspace of V.

E. vector v=(0,0,0) is the neutral element in V but it does not satisfy the conditions to be in set S. S here is not a subspace.

Here, is the set of twice differentiable functions with continuous first and second derivatives. The idea is basically the same as in A.F. I don't understand the notation for how they defined V.

ok.G. For the set of all polynomials of any degree f(x)=0 is the zero element. For all polynomials, f(0)=0. Set is is closed under addition but not multiplication. p(1)> 0 => p(0) but if you multiply the inequality by a negative scalar, the inequality doesn't hold. Not a subspace of V.