Here is a partial (negative answer) to your first question: 

**Proposition 1:** Two different vectors $(X_0,X_1)$ and $(X_0',X_1')$ cannot have
the exact same digits $d_0,d_1,\dots$ in base $(b_1,b_2)$, assuming $b_1,b_2>0$ and $b_1>b_2+1$. 

*Proof:* Suppose the contrary. Then for $k=0,1,\dots$ we have $X_{k+2}=b_1 X_k+b_2 X_{k+1}-d_k$, $X'_{k+2}=b_1 X'_k+b_2 X'_{k+1}-d_k$, and hence
$$Z_{k+2}=b_1 Z_k+b_2 Z_{k+1},$$
where $Z_k:=X'_k-X_k$. So, for some real $c_+,c_-$ and all $k=0,1,\dots$ we have 
$$Z_k=c_+ u_+^k+c_- u_-^k,$$
where 
$$u_+:=\frac{b_2+\sqrt{b_2^2+4b_1}}2,\quad u_-:=\frac{b_2-\sqrt{b_2^2+4b_1}}2$$
are the roots $u$ of the equation $u^2=b_1+b_2 u$. 

Note that $u_+>b_2\ge1$ and also $u_1>|u_2|$. So, if $c_+\ne0$, then $|Z_k|\to\infty$ (as $k\to\infty$), which contradicts the conditions $Z_k=X'_k-X_k$, $0\le X_k<1$, $0\le X'_k<1$. So, $c_+=0$. 

Now, for $b_2>0$, the condition $b_1>b_2+1$ is equivalent to $|u_-|>1$, whence $|Z_k|=|c_-|\,|u_-|^k\to\infty$ if $c_-\ne0$, which again contradicts the conditions $Z_k=X'_k-X_k$, $0\le X_k<1$, $0\le X'_k<1$. So, $c_-=0$, so that $Z_k=0$ and $X'_k=X_k$ for all $k$. In particular, $(X_0,X_1)=(X_0',X_1')$. $\Box$