The random variables $X,Y$ get only non-negative integer values and have common density function $f(m,n)=C\cdot (0.2)^m\cdot (0.3)^n$, for $m,n\geq 0$.
To calculate the value of $C$ we consider the property $$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(m,n)\, dm\, dn=1$$ ? And since $m,n\geq 0$ do we get $$\int_{0}^{+\infty}\int_{0}^{+\infty}f(m,n)\, dm\, dn=1$$ ?
In order to find the value of $C$ we evaluate the cdf
$$ f(m,n) = C \cdot (0.2)^{m} \cdot (0.3)^{n} | m,n \in \mathbb{N} \tag{1}$$
$$ \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} f(m ,n ) =1 \tag{2}$$
$$ \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} C \cdot (0.2)^{m} \cdot (0.3)^{n} \tag{3}$$ We can rewrite this. This is a geometric series. $$ \sum_{m=0}^{\infty} \frac{1}{5^{m}} = 1 + \frac{1}{5} + \frac{1}{25} + \cdots \tag{4}$$
The geometric series formula is $$ 1+ r + r^{2} + r^{3} + \cdots = \frac{1}{1-r} \tag{5}$$
$ \textrm{ Let } r =\frac{1}{5} \tag{6}$ $$ \frac{1}{1-\frac{1}{5}}= \frac{1}{\frac{4}{5}} = \frac{5}{4} = 1.25 \tag{7}$$
The following is the same as above
$$ \sum_{n=0}^{\infty} (\frac{3}{10})^{n} = 1 + \frac{3}{10} + \frac{9}{100} + \cdots \tag{8}$$ $$ \textrm{ Let } r = \frac{3}{10} \tag{9} $$ $$ \frac{1}{1-\frac{3}{10}} = \frac{1}{\frac{7}{10}} =\frac{10}{7} \tag{10}$$
Then we get
$$ \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} f(m,n) = C \frac{5}{4} \frac{10}{7} = C \cdot 1.\overline{7857142} \tag{11}$$
Then you see that $C$ is simply the reciprocal
$$ C = \frac{4}{5} \frac{7}{10} = \frac{28}{50} = \frac{14}{25}\tag{12}$$
Then we find that $f(m,n)$ is
$$ f(m,n) = \frac{14}{25}(0.2)^{m}(0.3)^{n} | m ,n \in \mathbb{N} \tag{13}$$
you can confirm with WolframAlpha