Calculating the hitting probability using the strong markov property ** This problem is from Markov Chains by Norris, exercise 1.5.4.**
A random sequence of non-negative integers $(F)n)_{n\ge0}$ is obtained by setting $F_0=0$ and $F_1=1$ and, once $F_0,\ldots,F_n$ are known, taking $F_{n+1}$ to be either the sum or the difference of $F_{n-1}$ and $F_n$, each with the probability $1/2$.  Is $(F_n)_{n\ge0}$ a Markov chain?
(a) By considering the Markov chain $X_n=(F_{n-1},F_n)$, find the probability that $(F_n)_{n\ge0}$ reaches $3$ before first returning to $0$.
(b) Draw enough of the flow diagram for $(X_n)_{n\ge0}$ to establish a general pattern.  Hence, using the strong Markov property, show that the hitting probability for $(1,1)$, starting from $(1,2)$, is $(3-\sqrt{5})/2$.
(c) Deduce that $(X_n)_{n\ge0}$ is transient.  Show that, moreover, with probability $1$, $F_n \rightarrow \infty$ as $n \rightarrow \infty$.

My attempt for (b): From $(1,2)$ the chain looks like $(1,2)\rightarrow (2,3)$ or $(1,2)\rightarrow (2,1)$ each with probability 1/2. From $(2,1)$ we can reach $(1,1)$. I want to calculate the probability generating function using the strong markov property $\phi(s)=\mathbb{E}_{(1,2)}(s^{H_{(1,2)}^{(1,1)}})$ where $H_{(1,2)}^{(1,1)}=\inf \{n\geq 0\colon X_n=(1,1) \text{ starting from } (1,2)\}$. I thought that if we start in $(2,3)$ and we want to reach $(1,1)$ we at least have to go trough $(1,2)$ again and then from $(1,2)$ to $(1,1)$. So I believe that $\mathbb{E}_{(2,3)}(s^{H_{(2,3)}^{(1,1)}})=\phi(s)^2$, but I am not sure if this true
I really need help with this exercise.
Thank you.

I'm self-studying this book recently. I'm not so sure about my answer. 
Part (a)
The probability is $\frac{3}{7}$.
Define $$h_{i,j} = \mathbb P(F_0 = i, F_1 = j, (F_n)_{n\geq 0} \textrm{ a Markov Chain reaches 3 before returning to 0})$$
Then all we need to figure out is:
$$ h_{0,1} = h_{1,1} = \frac{1}{2}h_{1,2}+\frac{1}{2}h_{1,0}$$
$$ = \frac{1}{4}h_{2,3}+\frac{1}{4}h_{2,1}+\frac{1}{2}h_{1,0}$$
$$ = \frac{1}{4}h_{2,3}+\frac{1}{8}h_{1,3}+\frac{1}{8}h_{1,1}+\frac{1}{2}h_{1,0}$$
$$ \Rightarrow \frac{7}{8}h_{1,1} = \frac{1}{4}h_{2,3}+\frac{1}{8}h_{1,3}+\frac{1}{2}h_{1,0}$$
$$ = \frac{1}{4}+\frac{1}{8}$$
$$ h_{0,1} = h_{1,1} = \frac{3}{7}$$
Part (b) by Dominik Scherempf is clear but I don't know why $h^{(1,2)}_{(2,3)}$ has the same distribution as $h^{(1,1)}_{(1,2)}$. Can we explain by the fact that they start from $(a,b)$ and end at $(b, a+b)$?
Part (c) to be solved. 
