= Solution
Linearizing about the orbit gives $\dot\eta=A(t)\eta$, where $A(t)=Df(X(t))$. The <fundamental matrix> therefore satisfies
$$
\boxed{\dot\Phi(t)=A(t)\Phi(t),\qquad\Phi(0)=I.}
$$
The <Floquet multipliers> are the <eigenvalues> of the <monodromy matrix> $\Phi(T)$. Differentiating the orbit equation shows that its tangent $\dot X(t)$ solves the <variational equation>. Since $\dot X(T)=\dot X(0)\ne0$, one multiplier is one.
For an invertible <fundamental matrix>, differentiating the <determinant> gives $(\det\Phi)'=\operatorname{tr}(\Phi^{-1}\dot\Phi)\det\Phi=\operatorname{tr}A\det\Phi$. <Integration> from the identity yields $\det\Phi(T)=\exp(\int_0^T\operatorname{tr}A(t)dt)$. The <determinant> is the product of the two multipliers, so the other is
$$
\boxed{\mu=\exp\left(\int_0^T\nabla\cdot f(X(t))\,dt\right).}
$$
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