Suppose A and B commute. Then let v be an eigenvector of A with eigenvalue \lambda. Then we have A(Bv)=(AB)v=BAv=\lambda Bv.
So Bv is in the eigenspace of A.
Choose a candidate basis {b_1,b_2,...,b_n} consisting of eigenvectors of A such that the eigenvectors are ordered to correspond with...
Well, if they gave you coordinates in terms of v1, v2, and v3, then you'd have a vector with 3 components, right? And any other basis for the same space would also have 3 components, because the dimension of the subspace is 3. So then you have your usual 3x3 thing for changing basis between those.
It might help if you gave an example of what you want to do. But I'm pretty sure the answer is yes, if it will help you solve the system (for example if the matrix has variable entries).
I don't really understand the question very well. "Changing basis" is a process of converting the representation of vectors in a space from one basis to another. But here you have two different spaces. It's kind of like asking "how do you change basis between \mathbb{R}^{3} and \mathbb{R}^{2}...
You'll have a hard time proving this statement because it's not true.
Consider the following matrix:
A= \begin{bmatrix}9 & 5.5 \\ 1 & 1\end{bmatrix}
The eigenvalues of A are both positive (verify).
Let x = (1,-3.1)T
Then
x^*Ax= \begin{bmatrix} 1 & -3.1...