Given x is an exponential random variable, find median & probability 
For the median, I believe that I should integrate the function, ∫x0λe−λtdt=1−e−λx
Then I need 1−e−λm=.5 for m, which is equivalent to e−λm=.5.
m=ln(2)/λ =>m=ln(2)/.2
 A: a) For the median, I believe your computation is correct, my only doubt is because of the lack of TeX. We want $m$ such that
$$\int_0^m \lambda e^{-\lambda x}\,dx=\frac{1}{2}.$$
Integrate. We want 
$$1-e^{-\lambda m}=\frac{1}{2},$$
or equivalently 
$$e^{-\lambda m}=\frac{1}{2}.$$
Take the logarithm of both sides. We get
$$-m\lambda=\ln(1/2)=-\ln 2,$$
and therefore $m=\frac{\ln 2}{\lambda}$. 
b) The probability that $X_1=X_2$ is $0$. (It is the integral over the line $x_1=x_2$ of the joint density function.)
By symmetry, $\Pr(X_1\gt X_2)=\Pr(X_2\gt X_1)$. By symmetry each is equal to $\frac{1}{2}$.
Or else we could integrate. The joint density function of $X_1$ and $X_2$ is $\lambda^2 e^{-\lambda x_1}e^{-\lambda x_2}$ for $x_1\gt 0$, $x_2\gt 0$, and $0$ elsewhere. Thus 
$$\Pr(X_2\gt X_1)=\int_{x_1=0}^\infty \left(\int_{x_2=x_1}^\infty \lambda^2 e^{-\lambda x_1}e^{-\lambda x_2}\,dx_2    \right)\,dx_1.$$
After some calculation we get $\frac{1}{2}$. Exploiting symmetry is a lot easier!
c) The probability that $Y\gt 3$ is equal to the probability both $X_1$ and $X_2$ are $\gt 3$. The probability that $X_1$ is greater than $3$ is $\int_3^\infty \lambda e^{-\lambda x_1}\,dx_1$. This is $e^{-3\lambda}$. For the probability that $Y\gt 3$, square. 
d) I do not know what tools you are expected to use, possibly the memorylessness of the exponential. Then we want the probability that an exponential with parameter $\lambda$ is $\gt 8$ given that it is $\ge 3$. This is the probability that an exponential with parameter $\lambda$ is $\gt 5$.
