First of all, a bit of intuition. The covariance of two random variables is a statistic that tells you how "correlated" two random variables are. If two random variables are independent, then their covariance is zero. If their covariance is nonzero, then the value gives you an indication of "how dependent they are".
Now, onto your problem.
I think you might find some use in the general formula for calculating expectations. For any random variable $X$ taking discrete values in $\mathbb{N}$, and any nonnegative measurable function $f$, you have
$$
\mathbb{E}[f(X)]=\sum_{k\in\mathbb{N}}f(k)\mathbb{P}(X=k).
$$
In your specific example, $Z=XY$ is a new random variable, and not a couple $(X,Y)$ as you seem to write in your example.
So, $Z$ is a random variable. What values does it take? Let us look at all the possibilities:
$$
\mathbb{P}(Z=1)=\mathbb{P}(X=1,Y=1)=0,
$$
because if $X=1$ then $Y=0$. Similarly,
$$
\mathbb{P}(Z=2)=\mathbb{P}(X=2,Y=1)=\mathbb{P}(X=2)=\frac16,
$$
because if $X=2$ then $Y=1$. Continuing the process, you find
$$
\mathbb{P}(Z=3)=\mathbb{P}(X=5)=0,\text{ and }\mathbb{P}(Z=4)=\mathbb{P}(Z=6)=\frac16.
$$
It remains to note that $Z$ can also take the value 0, and this happens whenever $Y=0$:
$$
\mathbb{P}(Z=0)=\mathbb{P}(X=1)+\mathbb{P}(X=3)+\mathbb{P}(X=5)=\frac16+\frac16+\frac16=\frac12.
$$
Once you know all the probabilities characterizing the distribution of $Z$, you can compute its expectation according to the previous formula:
$$
\mathbb{E}[Z]=2\cdot\frac16+4\cdot\frac16+6\cdot\frac16+0\cdot\frac12=\frac{12}6=2.
$$
Best Answer
1 - Not quite. The covariance (more specifically, the correlation coefficient), measures how much the variables "depend" on each other linearly. A great correlation coefficient shows us that one variable is "almost" a linear function of the other, I.e. there is $a$ and $b$ such that $Y \approx aX+b$. If two random variables are uncorrelated, that does not mean that here is no relation of dependence between them - just that this relation cannot be a linear one.
2 - Well, you just anwered this in your question: use the expectations! We can compute expectations of random variables even if they assume an uncountable number of values.