Do numbers higher than the base number have any meaning? For instance, in base 5, does this have any meaning 6 + 7 + 8 = ? or in base 2 2 + 3 + 4 = ?`
 A: Just to address the comment asking about whether it's mathematically sound.
It depends a bit on what you mean by mathematically sound. If you are writing some mathematical theorems or a paper where you for some strange reason have all numbers in base 5 then you shouldn't use 6,7 or 8 since it will be confusing. 
On the other hand the idea of bases is very much metamathematical in some sense. The numbers themselves have no idea of bases. The number 15 (in base 10) doesn't in itself know anything about the base it might be written in. Using bases to write numbers in positional notation is just something people came up with because typing $S(S(S(S(S(S(S(S(S(S(S(S(S(S(S(0)))))))))))))))$ is really tedious. And writing the set as a von Neuman ordinal would be even worse.
Edit Just a further clarification. I'm trying to distinguish here in some sense the situation of $5$ in $\mathbb{Z}_7$ and $5$ written in base $10$.
The first $5$ in some very real sense "knows" it's in $\mathbb{Z}_7$ as it's just short hand for $\{x;x \text{ mod } 7=5\}$. On the other hand the second $5$ usually knows nothing of the sort. We could of course think of the second $5$ as being short hand for $5*10^0$ but I don't think mathematicians usually do. They really just think of it as the "abstract" number $5$ possibly if you're closer to foundations and someone pushes you on the subject you might say it's $$\{\{\},\{\{\}\},\{\{\},\{\{\}\}\},\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\},\{\{\},\{\{\}\},\{\{\},\{\{\}\}\},\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}\}\}$$
What I would say is really mathematically unsound is writing $456_5$ as there we specify the base and yet have two digits of the wrong type. 
A: In base $5$, you only have $4$ digits: $0,1,2,3,4$. The digit $6$ does not exist in base $5$, instead, because $6=1\cdot 5^1 + 1\cdot 5^0$, the number $6$ is written as $11$ in base $5$. Similarly, $7$ is written as $12$ and $8$ as $13$.
You can still sum the numbers in any base, though, if you realise of course that you need to carry over $1$ if you hit a number over $5$, so
$$11+12+13 = 23 + 13 = 41$$
I got the $41$ because I first summed $3+3$, which is $11$ in base $5$, so I write down $1$ and carry over $1$, then I get $2+1=3$, and I add the carried over $1$, to get $4$.
