= Solution
{c}
A <product-Poisson stationary distribution> is suggested by balancing the three reaction fluxes around the directed cycle of <reaction complexes>. If their <mean> counts are $a,b$, this <complex balance> condition is
$$
2=\frac a2=ab,
$$
which gives $a=4$ and $b=1/2$. This suggests the <joint probability mass function>
$$
\pi(n,m)=e^{-9/2}\frac{4^n}{n!}\frac{(1/2)^m}{m!},\qquad n,m\geq0.
$$
It is normalized because it is the product of two <Poisson distributions>. We verify it directly in the <chemical master equation>, rather than assuming a <product-Poisson stationary distribution> theorem. The three incoming terms satisfy
$$
\begin{aligned}
2\pi(n-1,m)&=\frac n2\pi(n,m),\\
\frac n2\pi(n,m-1)&=nm\pi(n,m),\\
(n+1)(m+1)\pi(n+1,m+1)&=2\pi(n,m).
\end{aligned}
$$
With negative-index values set to zero, these identities also hold on the axes. Their sum equals $(2+n/2+nm)\pi(n,m)$, precisely the outgoing <probability> flux. The stationary average total exit rate is $2+\mathbb E_\pi n/2+\mathbb E_\pi(nm)=6<\infty$, so these <stationary balance equations> do define an <invariant distribution of a continuous-time Markov chain> for the nonexplosive model verified below. This is <complex balance>; it does not assert pairwise <detailed balance> for the irreversible <chemical reactions>.
To identify this <invariant distribution> with the limiting <probability distribution> in the question, first check that the model is a <nonexplosive continuous-time Markov chain>. On any finite time interval, $A_1$ is bounded by its initial count plus the finite number of arrivals in a rate-$2$ <Poisson process>. Conditional on that bound, the rate of producing $A_2$ is bounded, so only finitely many such productions occur. The number of joint removals cannot exceed the available molecules. Thus only finitely many <chemical reactions> occur in finite time.
The <continuous-time Markov chain> is also an <irreducible Markov chain>. From a <Markov chain state> with both counts positive, repeated joint removals reduce both. On the $A_2=0$ axis, producing one $A_2$ and then removing a pair reduces $A_1$ by one. On the $A_1=0$ axis, adding one $A_1$ and then removing a pair reduces $A_2$ by one. Each prescribed finite sequence has positive <probability>, so every <Markov chain state> reaches $(0,0)$. Conversely, arrivals followed by productions reach any $(n,m)$ with $n>0$; to reach $(0,m)$, reach $(1,m+1)$ and remove one pair. The <continuous-time Markov chain> is therefore <positive recurrent>, and the <continuous-time Markov chain convergence theorem> gives $\phi=\pi$ from the empty <initial condition>.
The two <Poisson> factors happen to have cancelling numerical coefficients at $(2,2)$:
$$
\boxed{\phi(2,2)=e^{-9/2}\frac{4^2}{2!}\frac{(1/2)^2}{2!}=e^{-9/2}.}
$$
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