 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 there are two pure subgame perfect equilibria, with strategiesThe 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, 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.