1) If $A$ is Hermitian positive definite, then $Ax=b$ is not ill-posed, since the matrix $A$ is invertible. It could be ill-conditioned if the condition of $A$, i.e., the quotient of its largest and smallest eigenvalue, is large.
2) No. For Hermitian matrices, eigenvectors are orthogonal, which is kind of an ideal situation. Ill-posedness is connected to the absolute values of eigenvalues.
If $A$ is only semi-definite (not positive definite), then $Ax=b$ is ill-posed. Tikhonov regularization means solution of $(A+\epsilon I)x=b$, which shifts the eigenvalues of $A+\epsilon I$ into the positive reals.