Let $X_0,X_1,...$ be a 3-state Markov chain with transition matrix
$$ \mathbf{P}=\begin{pmatrix} 0 & 1/2 & 1/2 \\ 1 & 0 & 0 \\ 1/3 & 1/3 & 1/3 \end{pmatrix} $$
and initial distribution $\alpha=(1/2,0,1/2)$. Find
(a) $\mathbb{P}(X_2=1|X_1=3)$
(b) $\mathbb{P}(X_1=3,X_2=1)$
(c) $\mathbb{P}(X_1=3|X_2=1)$
(d) $\mathbb{P}(X_9=1|X_1=3,X_4=1,X_7=2)$
The book I'm using only has answers for odd numbered assignments, this is an odd so I want someone to check my results. Here is what I've done:
(a) Because of time-homogenity of Markov chains we have that
$$\mathbb{P}(X_2=1|X_1=3)=\mathbb{P}(X_1=1|X_0=3)=P_{31}=1/3.$$
(b) This is a joint distribution and because of Markov independence property we have that
$$\mathbb{P}(X_1=3,X_2=1)=\mathbb{P}(X_2=1|X_1=3)\mathbb{P}(X_1=3)=\mathbb{P}(X_1=1|X_0=3)\mathbb{P}(X_1=3)=P_{31}(\alpha P)_3.$$
We have that
$$=P_{31}(\alpha P)=\frac{1}{3}\cdot \left(\frac{1}{2},0,\frac{1}{2}\right)\cdot \begin{pmatrix} 0 & 1/2 & 1/2 \\ 1 & 0 & 0 \\ 1/3 & 1/3 & 1/3 \end{pmatrix}=\left(\frac{2}{9},0,\frac{2}{9}\right),$$
so the sought probability is $2/9$.
(c) Since $X_1$ does not depend on $X_2$, we have that
$$\mathbb{P}(X_1=3|X_2=1) = \mathbb{P}(X_1=3)=(\alpha P)_3=\frac{1}{3}.$$
(d) We have that time-homogenity and Markov properties gives
$$\begin{align} \mathbb{P}(X_9=1|X_1=3,X_4=1,X_7=2)&= \mathbb{P}(X_9=1|X_7=2)\mathbb{P}(X_7=2|X_4=1)\mathbb{P}(X_4=1|X_1=3)\\ &\times\mathbb{P}(X_1=3)\\ &=\mathbb{P}(X_2=1|X_0=2)\mathbb{P}(X_3=2|X_0=1)\mathbb{P}(X_3=1|X_0=3)\\ &\times\mathbb{P}(X_1=3)\\ &=P^2_{21}P^3_{12}P^3_{31}(\alpha P)_{3} \end{align}$$
So we have that
\begin{align} P^2&=\begin{pmatrix}2/3 & 1/6 & 1/6 \\ 0 & 1/2 & 1/2 \\ 4/9 & 5/18 & 5/18 \end{pmatrix}\implies P^2_{21}=0,\\ \end{align}
so it does not matter what $P^3$ is. Thus the answer is $0$. If this answer is correct, what is the intuition behind it?
Your answer for (a) is correct.
Your reasoning for (b) is correct, but you made an error in computing $\alpha P$. It should be $\left(1/6,5/12,5/12\right)$, and so the desired probability is $1/3\cdot5/12=5/36$.
As @Did pointed out, $X_1$ and $X_2$ are not independent. Consider for example that on $\{X_2=1\}$, we have $\mathbb P(X_1=1)=0$. So for (c) we compute \begin{align} \mathbb P(X_1=3\mid X_2=1) &= \frac{\mathbb P(X_1=3,X_2=1)}{\mathbb P(X_2=1)}\\ &=\frac{\mathbb P(X_2=1\mid X_1=3)\mathbb P(X_1=3)}{\mathbb P(X_2=1)}\\ &=\frac{P_{31}(\alpha P)_3}{(\alpha P^2)_1}\\ &=\frac{1/3\cdot5/12}{5/9}\\ &=1/4. \end{align}
As @pwerth pointed out, the probability for (d) is simply $P_{31}^2=0$.