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