$X$ is an exponential random variable with mean $\beta=0.2$ and the event $A$ = {$X > 2$ }. What is $f_{X|A}(x)$?
The pdf of the exponential random variable is $f_{x}(x) = 0.2 e^{-0.2t}$ , $X >= 0$
Turning this into a CDF we get $F_{X}(x) = 1 - e^{-.2x}$, $X>=2)$
Using formula : $F_{X|A}(x) = F(x)/F(A) = \frac{F(\infty) - F(2)}{F(2)} = \frac{1 - (1 - e^{-0.2x})}{1 - e^{-0.4}} = \frac{e^{-0.2x}}{1 - e^{-0.4}} $
$f_{X|A}(x) =\frac{2e^{-0.2x}}{10(1 - e^{-0.4})}$, $X>=2$
I am unsure if I used the conditional pdf correctly in this question.
That is almost okay, but $A=\{X>2\}$ so $\mathsf P(A) {= 1-F_X(2)\\= 1-(1-\mathsf e^{-0.2\cdot 2})}$
Also, you do not need to differentiate a conditional CDF. You can get to the result directly by using the definition of a conditional probability density function over a given event: Also called "normalising".
$$\begin{split}f_{X\mid A}(x) &= \dfrac{f_X(x)~\mathbf 1_{x\in A}}{\mathsf P(A)} \\[1ex]& = \dfrac{f_X(x)~\mathbf 1_{x>2}}{1-F_X(2)} \\[1ex] &= \dfrac{0.2 \mathsf e^{-0.2x}\mathbf 1_{x>2}}{\mathsf e^{-0.2\cdot 2}} \\[1ex] &=0.2\mathsf e^{-0.2(x-2)}\mathbf 1_{x>2}\end{split}$$
An even faster way would be to invoke the memoryless property of exponential distributions ... unless you haven't proved that yet (and this does seem to be an exercise along the way towards that lesson).
So, anyway from this pdf, it is clear that: $X-2 \mid \{X>2\} \sim \mathcal{Exp}(0.2) $