Labelling the edges of a cube with {1, 2, 3,....,12} I  did the following problem:
a) Is it possible to label the edges of a cube by $1, 2, \cdots 12$ (using each number only once) so that at each vertex, the labels of the edges leaving that vertex have the same sum.
b) A suitable edge label  is replaced by $13$. Now is the equality of the sums possible?
The first is clearly not possible as, if each sum is equal to $k$ then we have, $78=4k$.
For the second one, to be possible (necessary condition) we get,
$78+13-i=4k$.
So, the possible values of $i$ are $3, 7, 11$.
Now my question is,
1)How to prove that for part b) the above condition is sufficient as well, without giving a construction?
I was able to give a construction(edge labelling) by some trial and error(smart guessing) for each $i=3, 7, 11$.
Or
2) Or is there a systematic way to give such a construction?
 A: If there exists an integer $i$ where $1 \le i \le 12$ such that $4 |( 91 - i)$ then equality of the sums is possible. $91$ is not divisible by $4$ but $92$ is. So perhaps we can rephase the above in the following manner to make things easier: let $j = i + 1$, if there exists an integer $j$ where $2 \le j \le 13$ such that $4 |( 92 - j)$, then equality of the sums is possible. Clearly $j$ must be a multiple of 4 so possible candidates are $4, 8, 12$. It then follows that $i$ can be $3, 7, 11$.  
EDIT: This is in response to the second part of your question where you were asking whether there's a systematic way to construct possible values of $i$.
A: One can write a program that finds the solutions, and yes there are solutions for all these.
Example of a python program that finds the solutions:
import itertools

edges = [
   (0, 1),      # 0
   (0, 2),      # 1
   (0, 4),      # 2
   (1, 3),      # 3
   (1, 5),      # 4
   (2, 3),      # 5
   (2, 6),      # 6
   (3, 7),      # 7
   (4, 5),      # 8
   (4, 6),      # 9
   (5, 7),      # 10
   (6, 7),      # 11
   ]

labels = [1,2,3,4,5,6,7,8,9,10,11,12]
labels[10-8] = 13 # 10 is the 11'th element
sref = sum(labels)/4 # this is the sum at each corner

cnt = 479001600

for x in itertools.permutations( labels ):
   cnt -= 1
   if cnt & 0xFFFFF == 0:
      print cnt

   if x[0] + x[1] + x[2] != sref:
      continue
   if x[0] + x[3] + x[4] != sref:
      continue
   if x[1] + x[5] + x[6] != sref:
      continue
   if x[3] + x[5] + x[7] != sref:
      continue
   if x[2] + x[8] + x[9] != sref:
      continue
   if x[4] + x[8] + x[10] != sref:
      continue
   if x[6] + x[9] + x[11] != sref:
      continue
   if x[7] + x[10] + x[11] != sref:
      continue

   print x

What can be noted is that there's $144$ solutions in the cases $i=3$ and $i=11$, and $48$ solutions in the case $i=7$. Since there's $24$ rotations of the cube and mirroring is possible the solution is unique except for this in the case $i=7$ and there's only three solutions otherwise.
