How to get the evolution equation of reduced density operator? Here is a simple system such that we can get the exact evolute equation for reduced density operator, but I'm not sure how to derive it from the result proved in the first part of this question. 
https://i.stack.imgur.com/93isy.jpg
EDIT: Someone has closed this question, so I come back to explain the main "concept" part underline the problem(though the question has already been solved). The key question is that "Will the evolution of reduced operator  be influenced by whether the state correspondents to the operator is mixed". 
 A: Intuitively, $H = H(1) + H(2)$ means that $\left\lbrace 1 \right\rbrace$ and $\left\lbrace 2 \right\rbrace$ evolve independently. If that is the case, you should be able to treat $\left\lbrace 1 \right\rbrace$ as follows:
$$\frac{d}{dt} \rho(1) = -\frac{i}{\hbar}\left[H(1), \rho(1) \right] \quad \quad \quad \mathbf{(1)}$$
To show this, start by writing $\rho(1) = \mathrm{Tr}_2(\rho)$, and use $\frac{d}{dt} \rho(1) = \frac{d}{dt} \left(\mathrm{Tr}_2(\rho) \right) = \mathrm{Tr}_2 \left(\frac{d}{dt} \rho \right)$.
From this, you have:
$$\frac{d}{dt} \rho(1) = \mathrm{Tr}_2 \left(\frac{d}{dt} \rho \right) = -\frac{i}{\hbar} \mathrm{Tr}_2 \left( \left[H(1) + H(2), \rho \right] \right) = -\frac{i}{\hbar} \mathrm{Tr}_2 \left( \left[H(1), \rho \right] \right) -\frac{i}{\hbar} \mathrm{Tr}_2 \left( \left[H(2), \rho \right] \right)$$
The first term is equal to what you expect (you can pass the trace over $\rho$ because $H(1)$ does not act on $\left\lbrace 2 \right\rbrace$).
The second term can be written as $-\frac{i}{\hbar} \left( \mathrm{Tr}_2 \left( H(2) \rho \right) - \mathrm{Tr}_2 \left( \rho H(2) \right) \right)$. Using the first part of the question, you can try to figure out why it is $0$.
As for physical interpretation, equation $\mathbf{(1)}$ shows that system $\left\lbrace 1 \right\rbrace$ is evolving "on its own", with the Hamiltonian $H(1)$, and is completely decoupled from system $\left\lbrace 2 \right\rbrace$.
