Extended game mixed SPNE I have doubts about whether my solution is right. 


Let $x&gt;z&gt;y&gt;1$.
Given the extended game above, we have two subgames and will find the Nash equilibria of these.

Observe that player 2 will play $b$ in the right subgame.
In the left subgame player 2 will be indifferent between playing $a$ and $b$.
Let $0 \le r \le 1$ such that $(r,1-r)$ is a mixed strategy for player 2 and such that the player plays this strategy in the left subgame. 

Player 1 has expected payoffs
$$
E u(s) = 
(rx+(1-r)y) \cdot 1_{s=A} + z \cdot 1_{s=B}
$$
Observe that player 1's best response is $A$ if $r \ge (z-1)/(x-y)$ and $B$ if $r \le (z-1)/(x-y)$, hence SPNE will be 
$$
\{(A,(r,1-r))|r \ge (z-1)/(x-y) \} \cup \{(B,(r,1-r))|r \le (z-1)/(x-y)\}
$$
Is this correct?
There are a couple of imprecisions. First, there are 3 subgames: two that start with player 2 moving, and the complete game is also a subgame. 

Second I think that A is a best response for player 1 iff $r\geq \frac{z-y}{x-y}$ otherwise, B is the best response (perhaps this was just a typo). 

Third, note that the fraction is guaranteed to be positive and smaller than 1 given the restrictions on the parameters,  but it is not necessarily equal to $0.5$, For example, let $x=5$, $z=3$ and $y=2$, then the fraction is equal to 1/3. 

Fourth, you are missing other equilibria, when $r=\frac{z-y}{x-y}$, then player 1 is indifferent between A and B, and any randomization of A and B also constitutes an equilibrium.

Lastly, when writing the equilibria, you must specify what player 2 does in each information set (or node in this game) 

Thus: $$NE=\{(A, (ra+(1-r)b)b)|r\geq \frac{z-y}{x-y}\}\cup\{(B, (ra+(1-r)b)b)|r\leq \frac{z-y}{x-y}\}$$
$$\cup \{((pA+(1-p)B), (ra+(1-r)b)b)|r=\frac{z-y}{x-y}, 0\leq p\leq 1\}$$

Where $(ra+(1-r)b)b$ specifies that player 2 randomizes in the first node and chooses b in the second node.