You’re not taking into account that the digit can occur in any of the six positions. There are $10\cdot36^5$ passwords that begin with a digit. There are also $10\cdot36^5$ passwords that end with a digit; some of these also begin with a digit and have already been counted, but some do not, so your figure of $10\cdot36^5$ is necessarily too small.
The easiest way to count the acceptable passwords is to note that there are $36^6$ six-character strings made up of upper-case letters and digits, and $26^6$ of them are made up entirely of letters, so there are $36^6-26^6$ that include at least one digit.
It is possible to count them directly, but the counting is more complicated. For each of the $6$ positions in the password there are $10\cdot36^5$ passwords having a digit in that position, so to a first approximation there are $6\cdot10\cdot36^5$ acceptable passwords. However, as noted in the first paragraph, this counts some passwords more than once. For each pair of positions in the password there are $10^2\cdot36^4$ passwords having digits in both of those positions, and all of these passwords have been counted twice. Since there are $\binom62$ pairs of positions, we must subtract $\binom62\cdot10^2\cdot36^4$ to get rid of the double-counting. Unfortunately, this overcompensates, and there are further corrections to be made. The net result, given by the inclusion-exclusion principle, is
$$\sum_{k=1}^6(-1)^{k+1}\binom6k10^k36^{6-k}\;.$$
Either way, the result is $1,867,866,560$.
Start with all $8$-character strings: $95^8$
Then remove all passwords with no lowercase ($69^8$), all passwords with no uppercase ($69^8$), all passwords with no digit ($85^8$) and all passwords with no special character ($62^8$).
But then you removed some passwords twice. You must add back all passwords with:
- no lowercase AND no uppercase: $43^8$
- no lowercase AND no digit: $59^8$
- no lowercase AND no special: $36^8$
- no uppercase AND no digit: $59^8$
- no uppercase AND no special: $36^8$
- no digit AND no special: $52^8$
But then you added back a few passwords too many times. For instance, an all-digit password was remove three times in the first step, then put back three times in the second step, so it must be removed again:
- only lowercase: $26^8$
- only uppercase: $26^8$
- only digits: $10^8$
- only special: $33^8$
Grand total: $95^8 - 69^8 - 69^8 - 85^8 - 62^8 + 43^8 + 59^8 + 36^8 + 59^8 + 36^8 + 52^8 - 26^8 - 26^8 - 10^8 - 33^8 = 3025989069143040 \approx 3.026\times10^{15}$
Best Answer
Hint: The number of passwords with exactly $2$ digits is $\binom{8}{2}(10^2)(52^6)$.
This is because the locations of the $2$ digits can be chosen in $\binom{8}{2}$ ways. For each of these ways, the $2$ chosen locations can be filled with digits in $10^2$ ways. And for each of these ways, the remaining $6$ slots can be filled with letters in $52^6$ ways.
Your calculation did not take into account the fact that there are quite a few choices for the location of the two digits.