= Solution
There are $kx_n$ seeds and their germinating proportion is $e^{-\gamma x_n}$, giving the <survival-augmented Ricker map> with zero survival. A positive <equilibrium point> satisfies $e^{-\gamma x_*}=1/k$, so
$$
x_*=\frac{\log k}{\gamma},\qquad F'(x_*)=1-\log k.
$$
It exists when $k>1$ and is strictly <linearly stable> when $|1-\log k|<1$. Thus \b[the usual strict linear-stability interval is] $\boxed{1<k<e^2}$.
With winter survival, surviving adults and new plants add:
$$
x_{n+1}=sx_n+kx_ne^{-\gamma x_n},\qquad x_* =\frac1\gamma\log\frac{k}{1-s}.
$$
Writing $L=\log[k/(1-s)]$, the <linearization of a dynamical system> has multiplier $1-(1-s)L$. Hence \b[the strict linear-stability interval becomes]
$$
\boxed{1-s<k<(1-s)\exp\!\left(\frac2{1-s}\right).}
$$
Its lower endpoint is less than $1$. Its upper endpoint exceeds $e^2$: for $t=1-s\in(0,1)$, $\log t+2/t$ is decreasing in $t$ and equals $2$ at $t=1$.
The <nonlinear stability at the Ricker flip threshold> needs a separate check. The upper endpoints have multiplier $-1$, so the strict linear test alone does not decide stability there. They are in fact locally <asymptotically stable>, with algebraic rather than geometric convergence. To see this, put $u=\gamma(x-x_*)$ at $L=2/(1-s)$. The local map is
$$
u\longmapsto-u+su^2+\left(\frac{1-s}{2}-\frac13\right)u^3+O(u^4),
$$
and its second iterate is $u-2(s^2-s/2+1/6)u^3+O(u^4)$. The cubic coefficient in parentheses is positive for every $s\in[0,1)$. Thus, if “stable” includes nonlinear endpoint stability, the upper inequalities can be replaced by $\leq$; the lower endpoints have no positive equilibrium.
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