Where each $X_i$ is independent. I know $E(X_i)=2$. So: $$E(Z) = E(2Y_1 + Y_2) = E[2\min(X_1,X_2) + \max(X_1,X_2)] = E\{2(X_1 or X_2) + (X_1 or X_2)\}$$
Since regardless of the outcome of the min and max functions we get the same expected values:
$$2E(X_i) + E(X_i) = 2(2) + 2 = 6$$
Only the book gives the answer as 5.
Note that $Y_1 + Y_2 = X_1 + X_2$ so $$E[Z] = E[X_1] + E[X_2] + E[Y_1]$$
The minimum of two independent exponential random variables with expected values $\theta_1$ and $\theta_2$ (and thus rates $1/\theta_1$ and $1/\theta_2$) is exponential with rate $1/\theta_1 + 1/\theta_2$ and thus expected value $\theta_1 \theta_2/(\theta_1 + \theta_2)$.
Since $Y_1+Y_2=X_1+X_2$ (why?), you get $$ \mathbb{E} Z = \mathbb{E}[X_1+X_2+Y_1]= \mathbb{E}[X_1]+\mathbb{E}[X_2]+\mathbb{E}[Y_1] = 4 + \mathbb{E}[Y_1] $$ so the question boils down to: "what is the expected value of the minimum of two iid exponential r.v.'s"? It is a standard result (or exercise) to show, e.g. with the cdfs, that the minimum of will also be exponential, with rate $\lambda = \frac{1}{\theta}+\frac{1}{\theta} = 1$ (i.e., $\theta^\prime= 1/\lambda =1$).
