We need to calculate cov(X,Y)=EXY - EX*EY, var(X) and var(Y). To do it, we have to know marginal distributions of both random variables X and Y. This can be done by "integrating the other variable out" of the joint density function. I will show you how to calculate the marginal density function of X:
$ f(x)= \int^{1}_{x} f(x,y) dy = \int^{1}_{x} 2 dy = 2(1-x)$ for $x \in (0,1)$.
Are you able to continue?
You can't. Here are two possible joint distributions which both yield the correct marginal distributions from your question, but which are different.
$$\begin{array}{l|l|l|}
& \mathbf{s}=s_1 & \mathbf{s}=s_2 \\
\hline
(\mathbf{u},\mathbf{v})=(x_1,x_1) & \frac{2}{6} & \frac{1}{6} \\
(\mathbf{u},\mathbf{v})=(x_1,x_2) & \frac{2}{6} & \frac{1}{6} \\
(\mathbf{u},\mathbf{v})=(x_2,x_1) & \frac{1}{6} & \frac{2}{6} \\
(\mathbf{u},\mathbf{v})=(x_2,x_2) & \frac{1}{6} & \frac{2}{6} \\
\end{array}$$
$$\begin{array}{l|l|l|}
& \mathbf{s}=s_1 & \mathbf{s}=s_2 \\
\hline
(\mathbf{u},\mathbf{v})=(x_1,x_1) & \frac{1}{6} & \frac{2}{6} \\
(\mathbf{u},\mathbf{v})=(x_1,x_2) & \frac{3}{6} & 0 \\
(\mathbf{u},\mathbf{v})=(x_2,x_1) & \frac{2}{6} & \frac{1}{6} \\
(\mathbf{u},\mathbf{v})=(x_2,x_2) & 0 & \frac{3}{6} \\
\end{array}$$
(Note that I renamed your random variable $\mathbf{x}_1$ to $\mathbf{u}$ and $\mathbf{x}_2$ to $\mathbf{v}$. Using the same name for random variable and for the elements of their image, i.e. $X = (x_1,x_2)$ seemed confusing)
Best Answer
If $Z = \max(X, Y)$, then the following is true:
$Z = 1$ iff $X = 1$ and $Y = 1$.
$Z = 2$ iff either $X = 1$ and $Y =2$ or $X = 2$ and $Y = 1$, or $X = 2$ and $Y = 2$.
$Z = 3$ iff $X = 3$.
So $P(Z = 1) = 0.12$ $P(Z = 2) = 0.08 + 0.12 + 0.18 = 0.38$ $P(Z = 3) = 0.2 + 0.3 = 0.5$.
Thus $E(Z) = 0.12 + 2*0.38 + 3*0.5 = 0.12 + 0.76 + 1.5 = 2.38$ $E(Z^2) = 0.12 + 4*0.38 + 9*0.5 = 0.12 + 1.52+ 4.5 = 6.04$
That means $V(Z) = E(Z^2) - (E(Z))^2 = 6.04 - 5.6644 = 0.3756$.