about possion gamma and exponential distribution about possion gamma and exponential distribution

can someone explain how to sub in the numbers? 
i tried subbing in numbers and the outcome its not the same as the answer. 
maybe I'm not doing it correct . 
I'm getting 1/2 e^-1.5 for the first question . 
 A: Exponential distributions deal with the amount of time between events.  The first problem is exponential with $\lambda = .5$. Since we are dealing with more than 3 hours between calls, we have
$$P(X\ge{3})=\int_{3}^{\infty}.5e^{-.5x}dx=-e^{-.5x}|_{3}^{\infty}=0-(-e^{-.5(3)})=e^{-1.5}=.2231$$
THe second problem, you need to know that the mean and standard deviation of an exponential distribution is $\frac1{\lambda}$.  Since this number is $2$, $(.5^{-1})$ then the we can find 
$$P(X\ge\mu+\sigma)=P(X\ge{4})=\int_{4}^{\infty}.5e^{-.5x}dx=-e^{-.5x}|_{4}^{\infty}=0-(-e^{-.5(4)})=e^{-2}=.1353$$
EDIT:  Continued answer.
For questions 3 and 4, we are talking about Poisson, since we are wanting the number of events given a mean and a particular interval.  Since we know the mean in one hour is .5, the mean in a 3 hour interval is 1.5.  Thus we will use $\lambda=1.5$.  Now we are looking for exactly 2 calls in three hours, thus, using Poisson,
$$P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!}=\frac{1.5^2e^{-1.5}}{2!}=.2510$$
Now for 4, we have our $\lambda=1$ since it is in a 2 hour interval.  The mean and variance for Poisson are both $\lambda$, so $\mu+2\sigma=1+2\sqrt1=3$.  Since we need fewer than 3 we have
$$P(X\le2)=\sum_{k=0}^{2}\frac{1^ke^{-1}}{k!}=e^{-1}\left(1+1+\frac1{2}\right)=.9197$$
