Is it possible that shortest path between two vertices changes scale all the edge-weights? Is it possible that shortest path between two vertices changes if we add ten to all the weights of all edges in the graph? How about if we multiply all the edges-weight by ten?
 A: Yes, if you add $10$ to the weights of all the edges. No if you multiply all edge-weights by $10$. 
Take the graph with vertices $a$, $b$, and $c$ with weighted edges $(a \to b, 1)$, $(b \to c,1)$, and $(a \to c, 3)$. The $a \to b \to c$ path is the shortest path from  $a$ to $c$ now, but if you add $10$ to each weight, the direct path $a \to c$ will be shorter. 
In the case of multiplication, if $N$ is the sum of the weights along the shortest path and $M$ is the sum of the weights of some competing path (so $N< M$), if you multiply all the edge-weights by $10$, that $10$ will factor out of the sum and you will still have $10N < 10M$.
A: Multiplication all weights by $10$ also changes weights of all pathes $10$ times, so the shortest one remains to be the shortest.
Addition is something else. Let $G \cong K_3$, $V(G) = \{\,1, 2, 3\,\}$ and $w(\{\,1, 2\,\}) = 1$, $w(\{\,1, 3\,\}) = 1$, $w(\{\,2, 3\,\}) = 3$. Then shortest $(1, 3)$-path is $1, 2, 3$, but after adding $10$ to all weights $1, 3$ becomes the shortest path.
