Find the number of all four-digit positive integers that are divisible by four and are formed by the digits 0,1,2,3,4,5
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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.
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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, $6^2-6^1 = 6(6-1)$.
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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.
It seems correct to me, though you may have made it more complicated than it needs to be.
The multiplication rule is quite appropriate here. We have four slots to fill; the first can be filled in five ways (since it can't be zero), the second can be filled in six ways, and the last two together can be filled in nine ways (as Arturo Magidin pointed out, 00 works as well). This gives us $5*6*9 = 270$ possibilities.