I was going through this question given to my friend at university, and I can do it with a little calculation. Is there a set algorithm that will work or a step-by-step guide that we can use so that you can solve this question for different value like $9$ and $10$?
Suppose the question asks for each side of the triangle summing up to $c$.
Then the sum of all three sides, counting each corner twice, is $3c$.
By this requirement, the sum of numbers in the corners is $3c-1-2-3-4-5-6 = 3c-21$.
For example, consider part (c), where $c = 11$.
Then the corner numbers sum to $3\times11-21=12$.
The three numbers, taken from $1$ to $6$, must be one of the triples: $$(3,4,5),(1,5,6),(2,4,6)$$
Which number belongs to which corner does not matter by symmetry.
By $3+4+6>11$ and $5+6=11$ we conclude the first two triples are invalid.
Thus the corner numbers must be $2, 4, 6$.
Finally it's a simple process to fill in the rest of the numbers.