For (i): let $\xi(0)$ be a small perturbation. Then $\xi(nP) = F(P)^n\,\xi(0)$. Decompose $\xi(0)$ into eigenspaces of $F(P)$; the component along $\dot{\hat{x}}(0)$ (eigenvalue $1$) corresponds to a phase shift that does not grow. The remaining components satisfy $|F(P)^n\,v| \leq \|F(P)\|^n |v|$, and since all non-unit eigenvalues have modulus strictly less than $1$, powers of $F(P)$ acting on the complementary subspace contract geometrically. Thus $|\xi(nP)| \to 0$ as $n \to \infty$ (in the direction transverse to the orbit), giving asymptotic stability. For (ii): the component of $\xi(0)$ along an eigenvector with $|\mu_i| > 1$ grows as $|\mu_i|^n$ after $n$ periods, so the perturbation grows without bound in that direction. This contradicts Lyapunov stability.