1.a. We must show the only linear combination of elements of A giving the 0 polynomial is the trivial combination. So, let c_0 p_0 + ... + c_n p_n = 0 for real numbers c_i. Plug in x=a_i. For each p_j with j != i, a_i is a root, so p_j(a_i) = 0. Thus c_i p_i(a_i) = 0. But p_i(a_i) is non-zero, so c_i = 0.
b. Yes. V is a vector space of dimension n+1 (standard basis: monomials 1=x^0, x^1, x^2, ..., x^n), and a linearly independent set of size equal to the dimension must be a basis.

2. If W=V, certainly the dimensions agree. On the other hand, if the dimensions agree, a basis for W, which exists, is necessarily a basis for V, so they are both given by the same linear combinations, hence W=V.

3.a. You want the matrix to take in coordinates of vectors in the first basis and spit out coordinates of vectors in the second basis. In particular, (1, 0) in the first basis means 2x-1, which is (-1, 2) in the second basis. Also, (0, 1) in the first basis means 2x+1 which is (1, 2) in the second basis. It follows that the matrix is
(-1 1)
(2 2)

[In general, the first column corresponds to the first standard coordinate vector's image, and likewise with the other columns.]

b. Nearly identical. (1, 0) means 1 = [(2x+1)-(2x-1)]/2 corresponding to (-1/2, 1/2). Similarly (0, 1) means x = ((2x-1)+(2x+1))/4 corresponding to (1/4, 1/4), so the matrix is
(-1/2 1/4)
(1/2 1/4)


(The finer points of the reasoning for 1b and 2 depending on exactly what definitions and propositions you're using. There are several logically equivalent approaches.)