Given independent random samples $(X_1,...,X_m)$ and $(Y_1,...,Y_n)$, respectively, from the following distributions of $X$ and $Y: X \sim \lambda_1 \exp (-\lambda_1 x) I(x >0)$, and $Y \sim \lambda_2 \exp (-\lambda_2 y) I(y >0)$. Consider the problem of testing the null hypothesis $H_0:\lambda_1=\lambda_2 $ against the alternative: $H_1:\lambda_1 \neq \lambda_2 $.
a. Formulate the underlying testing problem as that of testing one of the parameters in a multiparameter exponential family, having expressed the family explicitly in terms of all the parameters and the corresponding statistics.
b. Give the formula of the UMPU test at level $\alpha$, in its conditional form, involving these statistics.
c. Describe how this test can be stated equivalently as an unconditional test. What are the ultimate test statistic and the rejection region (in terms of a known distribution)?
Some examples of the similar question:
Say $X$ and $Y$ are independent. $X \sim Bin(m, p_1), Y\sim Bin(n , p_2)$.We want to test
$$H_0: p_1=p_2 \text{against} H_1: p_1>p_2,$$
Then $(X,Y)\sim f_{\theta_1,\theta_2} (x,y)=e^{\theta_1 T_1 + \theta_2 T_2-A(\theta_1,\theta_2 )}h(x,y)$
with $\theta_1=\log \frac{p_1 (1-p_2)}{p_2 (1-p_1)}, T_1=X, \theta_2=\log \frac{p_2}{1-p_2}, T_2=X+Y$
$$H_0: \theta_1=0 \text{ against } H_1: \theta_1>0.$$
The UMPU test
$ \phi_0 (T_1,T_2) = \begin{cases} 1 , \text{if} T_1 > c(T_2) \\ \psi(T_2) , \text{if} T_1=c(T_2) \\ 0, \text{if} T_1<c(T_2) \end{cases}$
where $c(T_2)$ and $\psi(T_2)$ are such that $E_{\theta_1=0} (\phi_0 (T_1, T_2) | T_2)=\alpha$.
Since the conditional distribution $T_1=X$ given $T_2=X+Y=t_2$ is
$$P(X=x | X+Y=t_2)=\frac{{m\choose x}{n \choose t_2-x} }{m+n \choose t_2}$$
$c(t_2)$ and $\psi(t_2)$ are such that
$$\sum\limits_{x > c(t_2)} {m\choose x} {n \choose t_2-x}+ \psi(t_2) {m \choose c(t_2) } {n \choose t_2-c(t_2)}={m+n \choose t_2}\alpha$$
Here I provided the final solution. Hope you can show how it can be obtained following the steps you are given.
Considering that (see here)
$$\frac{\lambda_1 \bar{X}}{\lambda_2 \bar{Y}}=\frac{\frac{2 \lambda_1 m \bar{X}}{2m}}{\frac{2 \lambda_2 m \bar{Y}}{2n}} \sim \frac{\frac{\chi^2_{2m}}{2m}}{\frac{\chi^2_{2n}}{2n}} \sim F_{2m,2n}$$
for $$\color{blue}{H_0: \frac{\lambda_1}{\lambda_2}= \theta}$$
you can use the test with the following test statistic: $$\color{blue}{T = \theta \frac {\bar{X}}{\bar{Y}}}$$
and critical region:
$$\color{blue}{C= \mathbb R \setminus (f_{(2m,2n), 1-\frac{\alpha}{2}}, f_{(2m,2n), \frac{\alpha}{2}})}$$
where $\alpha$ is the significance level of the test. Note that $f_{(2m,2n), 1-\frac{\alpha}{2}} = \frac{1}{f_{(2n,2m), \frac{\alpha}{2}}} $.