Hypothesis testing with exponential distribution I am stuck trying to solve the following problem

I can do the first part, but I am struggling to find the test in the second paragraph. So far I have:
$$
L(H_0, H_1) = 3^n\textrm{exp}\left(-2\sum_1^n x_i \right)
$$
which is decreasing in the statistic
$$
T(\mathbf{X}) = \sum_1^n X_i \sim \textrm{Gamma}(n, \theta) \sim \frac{1}{2\theta} \, \chi^2_{2n}
$$
I have tried to use tables for $\chi^2_n$ but I am finding difficulty since I seem to need lower points of the distribution, rather than upper points. 
 A: (Assuming all your working is correct)
If you can't use a stats package to get the lower tail quantiles (like calling qchisq in R), and you can't find chi-square tables that give lower quantiles (some do), you could use F tables.
Note that an F is a ratio of two independent chi-squared values on their df.
So if the denominator has infinite d.f. it will converge to 1 and you just get a chi-square on its d.f. (the numerator). So how does this help because you still only have upper tail values, right?
You flip it; the $1-a$ quantile of an $F_{m,n}$ is the $a$ quantile of an $F_{n,m}$. So have infinite numerator d.f. and let the denominator df. be the original d.f you're looking for. Now look up the upper tail quantile of that $F$. Invert and multiply by the d.f. and you have the lower tail quantile of a chisquare with that df.
compare the two in R:
 qchisq(.05,5)
[1] 1.145476
> 5/qf(.95,Inf,5)
[1] 1.145476

Doing it by hand from tables (in this example, with 5 df): 
- Look up .95 quantile of $F_{\infty,5}$ ... we get 4.3650 (e.g. see these tables)
-  5/4.3650 = 1.145475 is then the .05 quantile of a $\chi^2_5$, almost exactly what R gave using qchisq
