First verify that $\overline{g}_1 = \theta$ and $\overline{g}_2 = \beta$. We'll now find Subgame perfect equilibrium for all possible values of $(\theta, \beta)$ satisfying $\theta > \beta> 1$.
To do so, we first maximize player 2's payoff with respect to his contribution taking as given player 1's contribution:
\begin{eqnarray*} \max_{g_2 \geq 0} & \ \beta\ln (g_1+g_2)- g_2 \end{eqnarray*}
and we get the best response strategy of player 2 as a function of player 1's contribution:
\begin{eqnarray*} g_2 = \max(\beta - g_1, 0) \end{eqnarray*}
Next, solve player 1's payoff maximization problem taking as given player 2's strategy
\begin{eqnarray*} \max_{g_1 \geq 0} & \ \theta\ln (g_1+g_2)- g_1 \\ \text{s.t.} & \ g_2 = \max(\beta - g_1, 0)\end{eqnarray*}
and we get
\begin{eqnarray*} g_1^* = \begin{cases} 0 & \text{if } \theta < e\beta \\ \theta & \text{if } \theta \geq e\beta\end{cases} \end{eqnarray*}
Consequently, the contribution of player 2 in a subgame perfect outcome is
\begin{eqnarray*} g_2^* = \begin{cases} \beta & \text{if } \theta < e\beta \\ 0 & \text{if } \theta \geq e\beta\end{cases} \end{eqnarray*}
For the next one, the cost of player 2 is
\begin{eqnarray*} c_2(g_2) = \begin{cases} g_2 & \text{if } g_1 \geq \overline{g}_1 = \theta \\ \lambda g_2 & \text{if } g_1 < \overline{g}_1 = \theta\end{cases} \end{eqnarray*}
First verify that in this case $\overline{g}_1 = \theta$ and $\overline{g}_2 = \dfrac{\beta}{\lambda}$. We'll now find Subgame perfect equilibrium for all possible values of $(\theta, \beta, \lambda)$ satisfying $1 <\theta \leq \dfrac{\beta}{\lambda} < \beta$.
To do so, we first maximize player 2's payoff with respect to his contribution taking as given player 1's contribution:
\begin{eqnarray*} \max_{g_2 \geq 0} & \ \beta\ln (g_1+g_2)- c_2(g_2) \end{eqnarray*}
and we get the best response strategy of player 2 as a function of player 1's contribution:
\begin{eqnarray*} g_2 = \begin{cases} \dfrac{\beta}{\lambda} - g_1 & \text{if } g_1 < \overline{g}_1 = \theta \\ \max\left({\beta} - g_1, 0\right) & \text{if } g_1 \geq \overline{g}_1 = \theta \end{cases} \end{eqnarray*}
Next, solve player 1's payoff maximization problem taking as given player 2's strategy
\begin{eqnarray*} \max_{g_1 \geq 0} & \ \theta\ln (g_1+g_2)- g_1 \\ \text{s.t.} & \ g_2 = \begin{cases} \dfrac{\beta}{\lambda} - g_1 & \text{if } g_1 < \overline{g}_1 = \theta \\ \max\left({\beta} - g_1, 0\right) & \text{if } g_1 \geq \overline{g}_1 = \theta \end{cases}\end{eqnarray*}
and we get
\begin{eqnarray*} g_1^* = \begin{cases} 0 & \text{if } \lambda < e \\ \theta & \text{if } \lambda \geq e\end{cases} \end{eqnarray*}
Consequently, the contribution of player 2 in a subgame perfect outcome is
\begin{eqnarray*} g_2^* = \begin{cases} \dfrac{\beta}{\lambda} & \text{if } \lambda < e \\ \beta - \theta & \text{if } \lambda \geq e\end{cases} \end{eqnarray*}
