You made two errors:
- You did not account for the fact that there are only $9$ choices for the leading digit when neither number in the pair of consecutive equal even digits is in the thousands place.
- You have subtracted numbers in which there are three or more consecutive even digits more than once.
First, we observe that there are $9 \cdot 10 \cdot 10 \cdot 10 = 9000$ four-digit positive integers.
Let $A_1$ be the set of four-digit positive integers in which the thousands place and hundreds place contain equal even digits. Let $A_2$ be the set of four-digit positive integers in which the hundreds place and tens place contain equal even digits. Let $A_3$ be the set of four-digit positive integers in which the tens place and units place contain equal even digits. Then $A_1 \cup A_2 \cup A_3$ is the set of four-digit positive integers that do contain consecutive equal even digits. By the Inclusion-Exclusion Principle, the number of four-digit positive integers that do contain consecutive even digits is
$$|A_1 \cup A_2 \cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|$$
$|A_1|$: Since we cannot use zero, there are four ways of choosing the even digit that occupies both the thousands and hundreds place. There are ten choices for each of the remaining digits. Hence, $|A_1| = 4 \cdot 10 \cdot 10 = 400$.
$|A_2|$: Since we cannot use zero, the thousands place can be filled in nine ways. There are five ways to choose the even digit that fills both the hundreds and tens places. There are ten ways to fill the units place. Hence, $|A_2| = 9 \cdot 5 \cdot 10 = 450$.
$|A_3|$: We can fill the thousands place in nine ways and the hundreds place in ten ways. There are five ways to choose the even digit that occupies both the tens and units places. Hence, $|A_3| = 9 \cdot 10 \cdot 5 = 450$.
$|A_1 \cap A_2|$: Since we cannot use zero, there are four ways to choose the
even digit that occupies the thousands, hundreds, and tens places. There are ten ways to fill the units place. Hence, $|A_1 \cap A_2| = 4 \cdot 10 = 40$.
$|A_1 \cap A_3|$: There are four ways to choose the even digit that occupies both the thousands and hundreds places and five ways to choose the even digit that occupies both the tens and units places. Hence, $|A_1 \cap A_3| = 4 \cdot 5 = 20$.
$|A_2 \cap A_3|$: There are nine ways to fill the thousands place. There are five ways to choose the even digit that occupies the hundreds, tens, and units places. Hence, $|A_2 \cap A_3| = 9 \cdot 5 = 45$.
$|A_1 \cap A_2 \cap A_3|$: Since we cannot use zero, there are four ways to choose the even digit that occupies the thousands, hundreds, tens, and units places. Hence, $|A_1 \cap A_2 \cap A_3| = 4$.
Therefore, the number of four-digit positive integers that do contain consecutive even equal digits is
\begin{align*}
|A_1 \cup A_2 \cup A_3| & = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|\\
& = 400 + 450 + 450 - 40 - 20 - 45 + 4\\
& = 1199
\end{align*}
Therefore, the number of four-digit positive even integers that do not contain consecutive equal even digits is
$9000 - 1199 = 7801$.
How many six-digit positive integers contain exactly two even digits?
Remember that an integer $n$ is said to be even if there exists an integer $m$ such that $n = 2m$. An integer that is not even is said to be odd. Hence, there are five even digits: $$0, 2, 4, 6, 8$$ and five odd digits: $$1, 3, 5, 7, 9$$
Your strategy of breaking the problem into two cases is correct. Let's correct your calculations for those cases.
Case 1: The leading digit is even.
Since the leading cannot be zero, there are four ways to fill the leading digit. There are five ways to choose the position of the other even digit and five ways to fill that position with an even digit. There are five ways to fill each of the four remaining positions with an odd digit.
$$4 \binom{5}{1} \cdot 5 \cdot 5^4 = 4 \binom{5}{1} 5^5$$
Case 2: The leading digit is odd.
Two of the remaining five positions must be filled with even digits. There are $\binom{5}{2}$ ways to select the positions of the even digits and five ways to fill each of those positions with an even digit. There are five ways to fill each of the four other positions with an odd digit.
$$\binom{5}{2} \cdot 5^2 \cdot 5^4 = \binom{5}{2} 5^6$$
Total: Since the above cases are mutually exclusive and exhaustive, the number of six-digit positive integers that contain exactly two even digits is
$$4 \binom{5}{1} 5^5 + \binom{5}{2} 5^6$$
In how many five-digit positive integers are there digits that appear more than once?
Hint: Subtract the number of five-digit positive integers in which all the digits are distinct from the total number of five-digit positive integers.
The total number of five-digit positive integers is $$9 \cdot 10 \cdot 10 \cdot 10 \cdot 10 = 90000$$ since the leading digit can be chosen in nine ways (as $0$ is prohibited) and each of the remaining digits can be selected in $10$ ways. Alternatively, the largest five-digit number is $99999$ and the largest number with fewer than five digits is $9999$, so the number of five-digit positive integers is $$99999 - 9999 = 90000$$ The number of five-digit positive integers with distinct digits is $$9 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 27216$$ since we have nine choices for the leading digit (as $0$ is prohibited), nine choices for the thousands digit (since we cannot use the leading digit), eight choices for the hundreds digit (since we cannot use the tens thousands or thousands digits), seven choices for the tens digit, and six choices for the units digit. Therefore, the number of five-digit positive integers that do contain a digit that appears more than once is $$9 \cdot 10 \cdot 10 \cdot 10 \cdot 10 - 9 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 90000 - 27216 = 62784$$
Best Answer
If, as indicated in the edit, $0$ is not a natural number, that means the product of digits can't be zero and there are only $9$ possibilities for each digit. Consequently every $10$ in your calculation should be a $9$. This leads to an answer of $1377$.
A quick python script confirms that there are exactly $1377$ integers with this property, so the answer you were given is wrong. (The largest possible product of digits is $9^6=729^2$.)
Incidentally, if $0$ is treated as a natural number, the answer is not $2600$, despite two previous posts. This is because numbers where $af$ is not a square but $b$ or $c$ is $0$ still give a square product. There are $19$ choices for $b,c$ with at least one $0$. There are $90$ overall possibilities for $a,f$ and of these $26$ have $af$ being a square (nine with $a=f$, nine with $f=0$, six combinations of $1,4,9$ and two of $2,8$) so there are $19\times 64=1216$ cases missing from the other answers, giving a total of $3816$.