= Solution
Apply <backward induction in games>, or <Zermelo's algorithm>, from the terminal decision points. At node 4, player 1 chooses $R_4$ because $6>3$, giving continuation payoffs $(6,9)$. At node 5, player 1 chooses $L_5$ because $4>2$, giving $(4,6)$. At node 3, player 2 compares its continuation payoffs nine and six, so chooses $L_3$. At node 2, player 2 chooses $R_2$, receiving four rather than two.
At the root, player 1 consequently receives six from either action: $L_1$ leads to $(6,4)$ and $R_1$ to $(6,9)$. Thus \b[there are two pure <subgame perfect equilibria>], with strategies
$$
\boxed{(L_1,R_4,L_5;\ R_2,L_3),\qquad(R_1,R_4,L_5;\ R_2,L_3).}
$$
The semicolon separates the players, and each list specifies the player's action at every one of its decision points. All four non-root choices are strict. If mixed strategies are included, \b[every mixture between $L_1$ and $R_1$ at the root is also an equilibrium], with those other actions fixed. There are no further mixtures at the strict decision points.
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