It is easier to count the total directly: you have five possibilities for the first digit (any digit except $0$), and six each for the remaining three digits, so the total number is $5\times 6^3$. This agrees with your count ($6^4-6^3 = 6^3(6-1) = 6^3\times 5$), but I think it's easier to just count them directly.
You can use the same method for counting the possibilities of the first two digits: five possibilities for the first digit, six for the second, for a total of $5\times 6$; this is the same as your count, $6^2-6^1 = 6(6-1)$.
The rest is almost correct; again, you can be a bit briefer with the count of the last two digits: the first digit can be any of the six possibilities. If the first digit is 0, 2, or 4, then the second digit must be 0 or 4 to get a multiple of $4$; so for each of the three possibilities you have two $2$-digit combinations, giving $3\times 2=6$ possibilities. If the first digit is $1$, $3$, or $5$, then the second digit must be $2$, so you have only three more. The total here is the sum of the two, so we have $6+3=9$ possibilities.
What you forgot is that $00$ also works.
Since the choices for the first two digits are independent of the choices for the last two, so you multiply the totals. The total will then be
$$5\times 6 \times 9 = 270.$$
The fact that the total number (1080) is larger than the count of those that are divisible by 4 should not be a surprise: there are lots of numbers among the 1080 4-digit numbers in which every digit is one of 0, 1, 2, 3, 4, and 5 that are not divisible by $4$. So I'm not sure what your last line is meant to represent.
A three-digit positive integer has the form $100h + 10t + u$, where the hundreds digit $h$ satisfies the inequalities $1 \leq h \leq 9$, the tens digit $t$ satisfies the inequalities $0 \leq t \leq 9$, and the units digit
$u$ satisfies the inequalities $0 \leq u \leq 9$. Therefore, we wish to determine the number of solutions of the equation
$$h + t + u = 12 \tag{1}$$
subject to these restrictions.
If we let $h' = h - 1$, then $0 \leq h' \leq 8$. Substituting $h' + 1$ for $h$ in equation 1 yields
\begin{align*}
h' + 1 + t + u & = 12\\
h' + t + u & = 11 \tag{2}
\end{align*}
Equation 2 is an equation in the non-negative integers. A particular solution corresponds to the placement of two addition signs in a row of eleven ones. There are
$$\binom{11 + 2}{2} = \binom{13}{2}$$
such solutions since we must choose which two of the thirteen symbols (eleven ones and two addition signs) will be addition signs.
However, we have counted solutions in which $h' > 8$, $t > 9$, or $u > 9$. We must exclude these.
Suppose $h' > 8$. Then $h'$ is an integer satisfying $h' \geq 9$. Let $h'' = h' - 9$. Then $h'' \geq 0$. Substituting $h'' + 9$ for $h'$ in equation 2 yields
\begin{align*}
h'' + 9 + t + u & = 11\\
h'' + t + u & = 2 \tag{3}
\end{align*}
Equation 3 is an equation in the non-negative integers. Since a particular solution corresponds to the placement of two addition signs in a row of two ones, it has
$$\binom{2 + 2}{2} = \binom{4}{2}$$
solutions.
Suppose $t > 9$. Then $t$ is an integer satisfying $t \geq 10$. Let $t' = t - 10$. Substituting $t' + 10$ for $t$ in equation 2 yields
\begin{align*}
h' + t' + 10 + u & = 11\\
h' + t' + u & = 1 \tag{4}
\end{align*}
Equation 4 is an equation in the non-negative integers with $\binom{3}{2} = 3$ solutions, depending on which variable is equal to $1$. By symmetry, there are also three solutions in which $u' > 9$. No two of these restrictions cannot be violated simultaneously since $9 + 10 = 19 > 12$. Thus, the number of three-digit positive integers with digit sum $12$ is
$$\binom{13}{2} - \binom{4}{2} - 2\binom{3}{2} = 66$$
Best Answer
The interval $[10^r,10^{r+1}-1]$ contains all $(r+1)$-digit numbers. If a number in that interval consists of just $0$'s and $2$'s, then it must start with a $2$, followed by $r$ trailing digits. The number of such numbers is $2^r$, so the total number of trailing digits is $r2^r$; and by symmetry exactly half of the trailing digits are $0$, giving a total of $r2^{r-1}$ $0$'s.
Now just add up $r2^{r-1}$ for $0\le r\le 9$. (If you don't want to do this by hand, you will find tips for summing the series here, but doing it by hand is probably just as fast.)