Done

[limoz]

Completedcompletedcompletedcompletedcompletedcompletedcompletedcompletedcompletedcompletedcompletedc ompletedcompletedcompletedcompletedcompletedcompletedcompletedcompletedcompletedcompletedcompletedco mpletedcompletedcompletedcompletedcompletedcompletedcompletedcompletedcompletedcompletedcompleted

[opus]

-a
Regarding the last digits the consecutive numbers can only be 1 to 8 or 2 to 9.
The sum of both will in any case be a even number and the ending number 2 is always within the numbers. Even numbers do dived evenly by 2, one Housepoint!

[opus]

-b
22,23,24,25,26

[opus]

-c
The possible last digits are: 1 to 6 up to 4 to 9, all other sequences have a value that is a multiple of 10!
So there are allways 3 even and 3 odd numbers.

The possibls sums of those six number will be 21, 27, 33, 39 or those values plus multiples of 60.

The possible sums are allways odd, so no even number could fit, that makes a max of 3 housepoints, so far.
Within the possible sequneces is allways the digit 5, since no possible sum ends on a 5, we can't get a housepoint on that odd number, so the max housepoints are 2!
Done!
(Still thinking on -b)

[opus]

-b
We are looking for 5 consectuvie numbers, none of those numbers shall be a multiple of 10.
The possible sequences are 1 to 5, 2 to 6, 3 to 7, 4 to 8 and 5 to 9 and all the ones with added multiples of 10 (multiplicator*10)!
There sums will be 15, 20, 25, 30 or 35 or plus 50*multiplicator.
Since no odd sum could be divided evenly by a even ending digit, the sequences with 1 to 5, 3 to 7 and 5 to 9 can be neglected.
So we are looking for the least common multiple of the ending digits.
For 2 to 6 it would be 60, but 60 is no possible sum, the next one is 120. 120 is the sum of 22,23,24,25,26.
For 4 to 8 the least common multiple is 840, that is also no possible sum, so the first one in here is 1680.
So the solution is 22,23,24,25,26!