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Yes, similar matrices have the same eigenvalues.
Call your matrix A, the eigenvalues are defined as the numbers e for which a nonzero vector v exists such that Av = ev. This is equivalent to requiring (A-eI)v=0 to have a non zero solution v, where I is the identity matrix of the same dimensions as A. A matrix A-eI with this property is called singular and has a zero determinant. The determinant of A-eI is a polynomial in e, which has the eigenvalues of A as roots. Often setting this polynomial to zero and solving for e is the easiest way to compute the eigenvalues of A.
No, in general they do not. They have the same eigenvalues but not the same eigenvectors.
Yes. Simple example: a=(1 i) (-i 1) The eigenvalues of the Hermitean matrix a are 0 and 2 and the corresponding eigenvectors are (i -1) and (i 1). A Hermitean matrix always has real eigenvalues, but it can have complex eigenvectors.
Eigenvalues and eigenvectors are properties of a mathematical matrix.See related Wikipedia link for more details on what they are and some examples of how to use them for analysis.