(1) [EDIT: fixed significantly with respect to the first version] Let $\lambda$ be an eigenvalue of $A$, and $\tilde{\lambda}$ the closest eigenvalue of a perturbed matrix $A+E$. If $A=VDV^{-1}$ is diagonalizable, then $|\tilde{\lambda}-\lambda| \leq \kappa(V) \|E\|$. (Bauer-Fike theorem.)
If $A$ not diagonalizable, and $\lambda$ has multiplicity $k$, then $|\tilde{\lambda}-\lambda| = O(\|E\|^{1/k})$. In particular, this implies that the sensitivity is lower if the eigenvalues are distinct, since in one case the eigenvalues are locally Lipschitz and in the other they are not. However, $\kappa(V)$ may be very large; in particular, it has to diverge for a sequence of matrices that converge to a Jordan block, or a matrix with multiple eigenvalues. In this sense, matrices with multiple eigenvalues have worse sensitivity, but in practice there is little difference between a Jordan block and a small perturbation of it.
The additional problem is that you can't reliably tell those two cases apart, in floating-point arithmetic. For instance, the matrix $A = \begin{bmatrix}1 & -1/3 \\ 3 & -1
\end{bmatrix}$
has a Jordan block of size 2, but in any neighborhood of it there are infinite matrices with a cluster of two distinct eigenvalues. When one stores $A$ as a floating-point matrix the value $-1/3$ gets approximated, turning it into a nearby matrix $A+E$ with two distinct eigenvalues and $\kappa(V)\approx 3\cdot 10^{20}$. Even in cases when a matrix is exactly representable in floating point, the standard backward stable eigenvalue algorithms guarantee only that they compute the exact eigenvalues of a matrix $A + E$ with $\|E\| / \|A\| = O(u)$, where $u$ is the machine precision; so in presence of clusters they can't answer the question if eigenvalues are equal or not, and not even the question of checking if they are equal up to machine precision.
Knowing in advance that the multiplicity is high makes it an entirely different problem. "Compute the eigenvalues of a $2\times 2$ matrix $A$, knowing that it has a double eigenvalue" is a very simple problem; its condition number is 1, and it can be solved trivially by the algorithm lambda = trace(A) / 2
.
(2) Show us your experiments and maybe we might be able to tell you why.