Let me know if there are any errors in my logic
Find $a$:
We need to get the integral because we know if this is a real probability the sum of the integral from 0 to 3 will be 1.
$\int ae^{-x}$ from 0 to 3 which is $-ae^{-3}+ae^{0}=a1-ae^{-3}=a(1-e^{-3})$, which means a=$1/(1-e^{-3})$ for the integral to equal to 1
Find it's CDF:
Integral from 0 to x so. $(1/(1-e^{-3}))(1-e^{-x})$
Find $P(X \in [1,2])$:
So I'm guess it's the range so it wants [0,2]-[0,1]
[$(1/(1-e^{-3}))(1-e^{-2})$-$(1/(1-e^{-3}))(1-e^{0})$]-[$(1/(1-e^{-3}))(1-e^{-1})$-$(1/(1-e^{-3}))(1-e^{0})$]
I don't think there are any errors in your logic, but I think you can simplify your second step. To find $P(X\in [1,2])$, you just need to compute $-\dfrac{1}{1-e^{-3}} (1-e^{-2})+\dfrac{1}{1-e^{-3}} (1-e^{-1})$.