Let each of the numbers from $1$ up to $49$ be written on a ball, and let all these balls be contained in a box. From this box, we randomly draw exactly $6$ numbers (without putting them back, so we can get each number at most once). We call the largest one of these $6$ drawn numbers $M$.
I now want to figure out if this random variable $M$ has an expected value and/or a variance, and want to calculate them if they exist.
I know that the expected value $\mathbb E(M)$ (if it exists) is given by $\mathbb E(M) = \sum_{1 ≤ n ≤ 49} n p_n$, where $p_n = \mathbb P(M = n)$ is the probability of $M$ being the number $n$. Since this sum is finite because there are only finitely many combinations of these $6$ drawn numbers, I would deduce that $\mathbb E(M)$ does indeed exist. However, I can't really see an easy way to actually calculate $\mathbb E(M)$, considering that there are quite a lot of possible combinations, way too many to write them all down manually. Surely there is an elegant way to do this?
As for the variance, this is defined be $\text{Var}(M) = \mathbb E((M - E(M))^2)$. Can I already deduce here as well that the variance does exist because $M$ can only take finitely many values? And if I want to calculate the variance, I would need to know the value of $\mathbb E(X)$ beforehand, it seems.
Yes, in a finite case like this the expected value and variance exist; you don't need to prove that in detail.
You can calculate the expected value from the expression you wrote, but quite often, including in this case, it's easier to calculate it like this:
\begin{align} E(M)&=\sum_{m=0}^{48}P(M\gt m) \\ &= \sum_{m=0}^{48}\left(1-P(M\le m)\right) \\ &= \sum_{m=0}^{48}\left(1-\frac{\binom m6}{\binom{49}6}\right) \\ &= 49-\binom{49}6^{-1}\sum_{m=0}^{48}\binom m6 \\ &= 49-\binom{49}6^{-1}\binom{49}7 \\ &= 49-\frac{43}7 \\ &= \frac{300}7 \\ &\approx42.86\;. \end{align}
Alternatively, take $6$ red balls, $43$ black balls and $1$ blue ball, arrange them uniformly randomly in a circle, break the circle into a line at the blue ball (removing it), and consider the position of the last red ball in the line. By symmetry, the $7$ segments between the $7$ coloured balls have the same expected length, so the last red ball is an expected $\frac{43}7$ balls away from the end at $49$, yielding the above result.
The variance is slightly more involved but can also be calculated, using a slightly simplified form of the expression that you wrote, $\operatorname{Var(M)}=E(M^2)-E(M)^2$.