Recurrence of $\{0,1,2\}^n$ tuples that don't contain $2$ followed immediately by $0$ 
I'm doing part (a) and need some hints with it.
My approach is to divide members of $\{t_n\}$ into 2 sets:
$\bullet$ n-tuples start with $0$, i.e. $(0,\_ \ ,\_ \ ,\_ \ ,\_ \ ,\_ \ ,...)$: there are $t_{n-1}$ of those (i.e. we fill in the blanks with members of $\{t_{n-1}\}$).
$\bullet$ n-tuples with $0$ in position $p$, for $p = 2,3,...,n-1$, $\ $ i.e. $(...,\_ \ ,\_ \ , 0,\_ \ ,\_ \ ,\_ \ ,...)$: we want to fill the blanks with $\{t_{n-1}\}$, except those that have $2$ at position $(p-1)$, 
$\ $and there are $t_{n-2}$ of those. Since there are $(n-2)$ possible values for $p$, the count would be $(n-2)(t_{n-1} - t_{n-2})$.
I get stuck when it's time to sum up the above. I think they overlap, especially those in the second bullet point, but can't decide where. 
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Edit: I've now realized that my division of cases is wrong, thanks to the answers by Henning Makholm and Hagen von Eitzen (and also to the comment by another user, which for some reason was gone). The first of the sets is good, but the second one doesn't complement it. I'm thinking more about this problem and will edit my question again if needed.

The first of the sets is a good start, but why isn't your other set simply the valid $n$-tuples that start with $1$ or $2$?
To extend a tuple from the first class, you can stick either a $0$ or an $1$ in front of it.
To extend a tuple from the second class, you can stick $0$, $1$, or $2$ in front of it.
This gives you a coupled first-order recurrence between the number of tuples in each of the sets as $n$ increases.
