Reasoning for confidence interval Suppose 
$$X_1,\dots,X_{20} \sim f_X(x;\beta)$$
where
$$f_X(x;\beta) = \frac{1}{\beta} e^{-\frac{x}{\beta}},\quad x>0;\beta>0$$
It can shown that ("details omitted")
$$P(0.52 \bar{X} \leq \beta \leq 1.67 \bar{X}) = 0.99, \quad \forall\beta>0 \tag{1}$$
Thus the 99% confidence interval for  $\beta$ is
$$(0.52 \bar{X} , 1.67 \bar{X})$$
My question is how $(1)$ come about? 
My attempt to reason
By definition,
$$P(\text{confidence interval for } \beta) = 0.99$$
There exists test statistics $A = A(X_1,\dots,X_{20})$ and $B = B(X_1,\dots,X_{20})$  such that
$$P(A \leq \beta \leq B) = 0.99$$
I have found that the MLE of $\beta$ is $\hat \beta = \bar{X}$. This can be involved into the confidence interval as functions of $X_i's$. 
That is, there exists $a_1,a_2 \in \mathbb{R}$ such that
$$P(a_1 \bar{X} \leq \beta  \leq \bar{X} a_2) = 0.99$$
Since $\hat \beta = \bar{X}$,
$$P(a_1 \leq \frac{\beta}{\hat \beta} \leq a_2) = 0.99$$ 
Now we assign $a_1 = x_{0.001}$ and $a_2 = x_{0.99}$ because finding the probability of such bounds yields $0.99$, 
$$P(x_{0.001} \leq \frac{\beta}{\hat \beta} \leq x_{0.99}) = 0.99$$ 
Now we compute the CDF of $\dfrac{\beta}{\hat \beta}$ by finding the CDF of $\hat \beta$ then applying the transform, $ Y = \frac{1}{X}$.
Here is what I"ve established, is this the right direction? I'm skeptical because the functions $A$ and $B$ are not general enough..
 A: With this edit I am entirely replacing my answer.  Instead of addressing the reasoning in the question, I'll just answer the short question that follows the words "My question is".  You have an exponential distribution with expected value $\beta$.  To see that that is the expected value, integrate:
\begin{align}
& \int_0^\infty x f(x)\,dx = \int_0^\infty \frac x \beta e^{-x/\beta} \, dx = \beta \int_0^\infty \frac x \beta e^{-x/\beta}\,\frac{dx}\beta = \beta\int_0^\infty u e^{-u}\,du \\[8pt]
= {} & \underbrace{\beta \int u\,dv = \beta\left(uv -\int v\,du \right) }_{\text{integration by parts}} = \beta\left( \left.\vphantom{\frac11}u e^{-u}\right|_0^\infty - \int_0^\infty -e^{-u}\,du  \right) = \beta(0+1).
\end{align}
The "$0$" at the end can be found by L'Hopital's rule.  There is also a common-sense way to see that it is $0$.  That the last integral is $1$ I'll leave as an exercise unless further questions are forthcoming.
Here's an exercise: (1) Show that $\beta$ is a scale parameter; (2) use that to show that the expected value must be $\beta$ times some constant, without finding any integrals.
From part (2) of the exercise, one sees that the point of evaluating the integral is just to show that the "constant" is $1$.
Similarly to part (2) of the exercise, one can show without finding any integrals that the standard deviation is some constant times $\beta$.  But let's find the variance by brute force.  First, the expected value of the square of this exponentially distributed random variable:
$$
\int_0^\infty x^2 f(x)\,dx = \int_0^\infty \frac{x^2}\beta e^{-x/\beta}\,dx = \beta^2 \int_0^\infty \left(\frac x \beta \right)^2 e^{-x/\beta}\, \frac{dx}\beta = \underbrace{\beta^2 \int_0^\infty u^2 e^{-u}\,du =2\beta^2}_{\text{Integrate by parts again.}}.
$$
Then
$$
\text{variance}=\text{expected value of the square minus the square of the expected value} = 2\beta^2-\beta^2.
$$
So the standard deviation is $\beta$.
Since $X_1,\ldots,X_{20}$ are independent, we get $\operatorname{var}(X_1+\cdots+X_{20}) = 20\beta^2$, so $\operatorname{var}\left(\dfrac{X_1+\cdots+X_{20}}{20}\right) = \dfrac 1 {20^2}\cdot 20\beta^2 = \beta^2/20$. And we don't need independence to show that $\operatorname{E}(\bar X) = \beta$.
Consequently $\bar X$ has expected value $\beta$ and standard deviation $\beta/\sqrt{20}$.  And so
$$
\frac{\bar X - \beta}{\beta/\sqrt{20}} \tag{$*$}
$$
has expected value $0$ and variance $1$.
Probably what was done next is that the central limit theorem was invoked and $(*)$ was treated as approximately normally distributed.  Hence its probability of being $<-2.5758$ is about $0.005$ and its probability of being $>2.5758$ is about $0.005$.  Thus the event
$$
-2.5758 < \sqrt{20}\frac{\bar X-\beta}{\beta} < 2.5758
$$
has probability about $0.99$.  Multiplying the numerator and denominator by $1/\beta$ and dividing both sides by $\sqrt{20}$ we get
$$
\frac{-2.5758}{\sqrt{20}} < \frac{\bar X}\beta - 1 < \frac{2.5758}{\sqrt{20}}
$$
$$
1-\frac{2.5758}{\sqrt{20}} < \frac{\bar X}\beta < 1+ \frac{2.5758}{\sqrt{20}}
$$
$$
\frac 1 {1-\frac{2.5758}{\sqrt{20}}} > \frac \beta {\bar X} > \frac 1 {1+ \frac{2.5758}{\sqrt{20}}} 
$$
$$
2.35834 > \frac \beta {\bar X} > 0.6345287
$$
$$
2.35834 \bar X >\beta > 0.6345287 \bar X
$$
This differs from what you had.  A possibility that I haven't checked is that making the lower bound smaller and also making the upper bound smaller will leave the probablity at $0.99$.  It is conceivable that that would be done in order to get a shorter interval.  Another possibility is that your arithmetic or mine has errors.
