Yes, this is a homework question. If I were designing this assignment, I would be using this question to show that a matrix could be diagonalizable in one field, but not another, so I was thinking that it would be NO for part (a) and YES for part (b). But I got no for both parts. Did I make a mistake? I'm confident on my answers for all of the rest of the assignment, but I'm concerned about this one. Thanks!
EDIT: fixed a sign error
You messed up the quadratic formula. It should be $\sqrt{(-2)^2-8}=2i$ rather than $\sqrt{(-2)^2-4(2)(2)}=2i\sqrt{3}$. So your eigenvalues should be $1\pm i$.
You can tell you made a mistake when you found that $A-(1+i\sqrt{3})$ was invertible, since the definition of an eigenvalue is that $\lambda$ is an eigenvalue if and only if $\det(A-\lambda)=0$.
This doesn't change the answer to the second part, but now you should be able to solve the second part using the correct eigenvalues. However, you don't even need to find the explicit vectors. Eigenvectors for distinct eigenvalues are linearly independent, so since $A$ has two distinct eigenvalues, it has two linearly independent eigenvectors. I.e. a basis of eigenvectors. Hence $A$ is diagonalizable over $\Bbb{C}$.