The joint probability generating function of $X$ and $Y$ is $G_{X,Y}(s,t) = E[s^X t^Y] = \sum_{x=0}^{\infty} \sum_{y = 0}^{\infty} p(x,y) s^x t^y.$
For $E[X]$, we have $E[X] = \left.\frac{\partial G_{X,Y}(s,t)}{\partial s} \right|_{s=1,t=1}.$ This is because taking the partial derivative of $G_{X,Y}(s,t)$ with respect to $s$ yields $\sum_{x=1}^{\infty} \sum_{y = 0}^{\infty} x p(x,y) s^{x-1} t^y.$ Then, subbing in $s=1, t = 1$ gives $\sum_{x=1}^{\infty} \sum_{y = 0}^{\infty} x p(x,y),$ which is how you calculate $E[X]$ directly from the joint distribution of $X$ and $Y$.
Similarly, $\begin{align}E[Y] &= \left.\frac{\partial G_{X,Y}(s,t)}{\partial t} \right|_{s=1,t=1}, \\ E[XY] &= \left.\frac{\partial^2}{\partial s \partial t} G_{X,Y}(s,t) \right|_{s=1,t=1}. \\ \end{align}$
To find $\text{Cov}(X,Y)$, remember that $\text{Cov}(X,Y) = E[XY] - E[X]E[Y].$ Then use the formulas you already know for $E[X], E[Y]$, and $E[XY]$.