Markov Chains: Example to show past and future are not independent given any information about past. The Markov property does not imply that the past and the future are independent given any information concerning the present. Find an example of a homogeneous Markov chain $\left\{X_n\right\}_{n\ge0}$ with state space $E=\left\{1,2,3,4,5,6\right\}$ such that 
$P(X_2=6\mid X_1\in\{3,4\},X_0=2)\neq P(X_2=6\mid X_1\in\{3,4\})$.
I don't understand the phrasing of this question. How should I approach this?
Could this work? 
$P(X_2=6\mid X_1\in\{3,4\}) = 1$
But $P(X_2=6\mid X_1\in\{3,4\},X_0=2) =0$

Suppose that $X_0$ is equally likely to be in states $1$ and $2$, each with probability $\frac{1}{2}$.
Suppose that from state $3$, there is a $100\%$ chance of going to state $6$ in the next step.
Suppose that from state $4$, there is a $100\%$ chance of going to state $5$ in the next step. 
Suppose that from state $1$, there is a $100\%$ chance of going to state $3$. 
Suppose that from state $2$, there is a $100\%$ chance of going to state $4$.

Now, the LHS of your equation will evaluate to $0$, because given that you were in state $2$ at $X_0$, you must go to state $4$ in $X_1$ and therefore state $5$ in $X_2$, which means there's a $0\%$ chance of being in state $6$ at $X_2$. 
The RHS of your equation will evaluate to $\frac{1}{2}$, because we are not conditioning on any value of $X_0$, so we use its initial probabilities: $0.5$ chance of $X_0=1$ and $0.5$ chance of $X_0=2$. This means that there's a $0.5$ chance of $X_3=6$ and $0.5$ chance of $X_3=5$. 

It will probably help you to draw the directed graph representing the Markov chain's transition probabilities. For good practice, also try to write out the transition matrix (it will be a $6\times 6$ matrix).
