Bertrand game with delivery risk and side payments Consider three agents $A_i$, who engage in a Bertrand game. All agents have perfect knowledge on all parameters and the distribution $F()$.

$A_1$ moves first and selects price $0\leq p_1\in \mathbb{R}$. $A_2$ moves second and selects price $0\leq p_2\in \mathbb{R}$. $A_3$ moves third and selects $0\leq q_1\in \mathbb{R}$ and $0\leq q_2\in \mathbb{R}$, that is, the quantities that it buys from $A_1$ and $A_2$, which it sells at market clearing price.
$A_1$ tries to maximize $  q_1 p_1$.
$A_2$ tries to maximize $ q_2 p_2$.
$A_3$ tries to maximize $$((A - q_1 - q_2) (q_1+q_2) - q_1 p_1 - q_2 p_2) (1-F(-q_1 p_1))+
( (A - q_2) q_2- q_2 p_2+B) F(- q_1 p_1) $$, with $F$ being a continuous cumulative distribution function. $A_3$'s profit function captures that buying from $A_2$ comes at no risk, while $A_1$ might be unable to deliver. However, the higher $A_1$'s profits the smaller that risk and the higher the chance to capture the benefits/losses $B$.

Note, all agents have perfect knowledge on all parameters and the distribution $F()$.
Does $A_3$'s profit increase in $B$? How to prove this?
(If $F$ is uniform, $A_3$ will only buy from one of the agents and it is easy to show that the answer is yes. If $p_1$ and $p_2$, then it is easy to show that the answer is yes.)
In the general case, which is outlined above, my numerical studies suggest that the answer is yes, but I struggle to write down the analytical proof.
I am not going to give a solution to you question (as I am not able too). However, let me try to indicate the difficulty of obtaining one if you cannot find a closed form solution for $q_1$ and $q_2$ in the last stage of the game.
Let $\Pi(q_1, q_2, p_1, p_2, B)$ be the objective function that we want to maximize. Assuming we can totally differentiate this, we get:
$$
d\Pi = \Pi_{q_1} dq_1 + \Pi_{q_2} dq_2 + \Pi_{p_1} dp_1 + \Pi_{p_2} dp_2 + \Pi_B dB. \tag{1}
$$
Assume the first order conditions hold:
$$
\Pi_{q_1} = 0\\
\Pi_{q_2} = 0 \tag{foc-stage3}
$$
This then simplifies $(1)$ to:
$$
d\Pi = \Pi_{p_1} dp_1 + \Pi_{p_2} dp_2 + \Pi_B dB \tag{1.a}
$$
The problem here is that we can not drop the $dp_1$ and $dp_2$ terms as they are (in general) both going to be a function of $B$.
Assume that the first order conditions $\rm{(foc-stage3)}$ can be solved to give a solution $q_2(p_1, p_2, B)$ and $q_1(p_1, p_2, B)$. Then in stage 2 of the game, Firm 2 maximizes:
$$
p_2 q_2(p_1, p_2, B)
$$
Again, assume that the first order conditions hold:
$$
q_2 + p_2 \frac{\partial q_2}{\partial p_2} = 0 \tag{foc-stage2}
$$
We can totally differentiate this with respect to $p_1$ and $p_2$ and $B$ (notice that $q_2$ is a function of all three these variables). This gives:
$$
\left(2\frac{\partial q_2}{\partial p_2} + p_2 \frac{\partial^2 q_2}{\partial p_2\partial p_2}\right) dp_2 = -\left(\frac{\partial q_2}{\partial p_1} + p_2 \frac{\partial^2 q_2}{\partial p_2 \partial p_1}\right) dp_1 - \left(\frac{\partial q_2}{\partial B} + p_2 \frac{\partial^2 q_2}{\partial p_2 \partial B}\right) dB \tag{2}
$$
Let $p_2(p_1, B)$ be the optimal solution of firm 2 (by solving $({\rm foc-stage2})$. Then in stage 1, Firm 1 maximizes
$$
p_1 q_1(p_1, p_2(p_1, B), B)
$$
So:
$$
q_1 + p_1 \frac{\partial q_1}{\partial p_1} + p_1 \frac{\partial q_1}{\partial p_2} \frac{\partial p_2}{\partial p_1}= 0 \tag{foc-stage1}
$$
Assuming again that this can be totally differentiated with respect to $p_1$ and $B$, we obtain:
$$
\begin{align*}
\left(\frac{\partial q_1}{\partial p_1} + \frac{\partial q_1}{\partial p_1} + p_1 \frac{\partial^2 q_1}{\partial p_1 \partial p_1} + \frac{\partial^2 q_1}{\partial p_2\partial p_1}\frac{\partial p_2}{\partial p_1} + \frac{\partial q_1}{\partial p_2}\frac{\partial^2 p_2}{\partial p_1\partial p_1}\right) dp_1\\
=- \left(\frac{\partial q_1}{\partial B} + p_1 \frac{\partial^2 q}{\partial p_1 \partial B} + \frac{\partial^2 q_1}{\partial p_2 \partial B} \frac{\partial p_2}{\partial p_1} + \frac{\partial q_1}{\partial p_2} \frac{\partial^2 p_2}{\partial p_1 \partial B}\right) dB \tag{3}
\end{align*}
$$
Then we can use $(3)$ to substitute for $d p_1$ in $(2)$. Then $(2)$ and $(3)$ give $dp_1$ and $dp_2$ in terms of only  $dB$. This can be substituted into $(1.a)$ to give the total effect of $B$ on $\Pi$.