Let $X\sim\mathrm{Exp}(1)$ and $Y\sim\mathrm{Exp}(2)$ independent random variables. Let $Z = \max(X, Y)$. Calculate expected value of $Z$. Here a question:

Let $X \sim \mathrm{Exp}(1)$ and $Y\sim\mathrm{Exp}(2)$ be independent random variables. Let $Z = \max(X, Y)$. Calculate expected value of $Z$.

My try:
\begin{align}
P(Z \le z) & = P(\max(X, Y) \le z) = 1 - P(\max(X, Y) > z) \\[10pt]
& = 1-P(X > z)P(Y > z) =  1 - (e^{-z} \cdot e^{-2z}) = 1 - e^{-3z}.
\end{align}

find distribution function: $F_z = \frac{d}{dz} \space 1 - e^{-3z} = 3e^{-3z}. $
Find expected value: $\int_0^\infty z\cdot F_z \,dz = 3\int_0^\infty z\cdot e^{-3z}\,dz = \frac{1}{3}$.

Right solution is $\frac{7}{6}$. Whats wrong in my way?
 A: We have $\Pr(Z\le z)=\Pr((X\le z)\cap (Y\le z))=(1-e^{-z})(1-e^{-2z})$ (for $z\gt 0$.) Differentiate to find the density, and then calculate the expectation as usual.
A: If the maxiumum of two variables is less than( or equal to) a constant, then both variables are.  
If the maximum of two variable is greater than a constant, then at least one variable is.$$P(\max(X,Y)\leq z)=P(X\leq z\cap Y\leq z)\\P(\max(X,Y)> z)=P(X>z\color{red}{\cup} Y>z)$$
Use the former.
A: You need
$P(Z \le z) = P(Max(X, Y) \le z) = P(X \le z)(Y \le z) = \cdots.$
Addendum: With a little thought one can find the mean of the max of two exponentials of equal rate $\lambda$ without finding the distribution. The minimum is exponential with rate $2\lambda,$
and thus mean $1/2\lambda.$ Then (invoking the no-memory property
of exponential distributions) the additional wait for the max is
exponential with rate $\lambda$ and mean $1/\lambda.$ Thus the
expectation of the max is $1/2\lambda + 1/\lambda = 3/2\lambda.$
This method can be easily extended to find the expectation of
the max of several independent exponentials with a homogeneous rate.
With a little extra though one can find the expectation of the max
of two exponentials $X$ and $Y$ with different rates, say $\lambda$ and $\mu,$ respectively.
The rate of the min is $\nu = \lambda + \mu$ and so its mean is
$1/\nu.$ To finish the wait, you need to know which exponential
waiting time came first. This is $X$ with probability $\lambda/\nu$ and $Y$ with probability $\mu/\nu.$ Then the expected wait from the min to the max is $(1/\mu)(\lambda/\nu) + (1/\lambda)(\mu/\nu).$
Then the total expected wait for the max is
$$(1 + \lambda/\mu + \mu/\lambda)/\nu.$$
For $\lambda = \mu$ it is easy to see that this is the same result
as above for two equal rates.
The brief bit of R code below simulates your problem and illustrates this approach for 
$\lambda = 1$ and $\mu = 2.$ (With a million cases, simulated results are accurate to about three places.)
 m = 10^6;  mu = 1;  lam = 2
 x = rexp(m,mu); y = rexp(m,lam)
 v = pmin(x,y);  w = pmax(x,y)
 mean(x == v)
 ## 0.333699      # approx. P(X first) = 1/3
 nu = mu + lam;  (1 + lam/mu + mu/lam)/nu
 ## 1.166667      # exact E(max)
 mean(w)
 1.166280         # approx. E(max)

A: $X$ and $Y$ are exponential random variables with parameters
$1$ and $2$ respectively, and so 
$$E[X] = 1,~ E[Y] = \frac 12.$$
$\min(X,Y)$ is an exponential random variable with parameter $3$ (cf. Gregory Grant's comment that $P\{\min(X,Y) > z\} = P\{X > z\}P\{Y > z\}$
and in your own calculations shown in your question, you found
that $P\{X > z\}P\{Y > z\} = e^{-z}\cdot e^{-2z} = e^{-3z}$.)  So, $$E[\min(X,Y)] = \frac 13.$$
 Now note that $\max(X,Y) + \min(X,Y) = X+Y$ and so
$$E[\max(X,Y)] = E[X] + E[Y] - E[\min(X,Y)
= 1 + \frac 12 - \frac 13 = \frac 76.$$
A: $P(\min(X,Y)>z)=P(X>z)P(Y>z)$.  Not the max.
A: Your line $P(Z\le z)=1-P(X>z)P(Y>z)$ is wrong. It excludes cases like $X=z-1, Y=z+1$. Some further thought however will show that $P(Z\le z)=P(X\le z)P(Y\le z)$ covers all the cases and will give you the right answer.
A: Let $g(x,y) = 2 e^{-x} e^{-2y}.$  Then 
$$
P(\max(X,Y))= \int^\infty_0 \int^x_0 x \,g(x,y) \ dy \ dx +  \int^\infty_0 \int^\infty_x y \, g(x,y) \ dy \ dx =\frac{8}{9} + \frac{5}{18}  = \frac{7}{6}.
$$
