Finding the two shortest paths while minimizing the number of nearby/common edges The shortest path problem between 2 arbitrary nodes is one that has been covered extensively and the solution is well-known. Consider the edge costs to be arbitrary.

Consider the following variant:

Find the two shortest paths between 2 pairs of arbitrary nodes, however  the cost is considered to be the length of the path and in addition the number of nearby/common edges in the two paths. This cost is $|P_1| + |P_2| + |P_1 \cap P_2|$ when considering only common edges. More generally we can state the cost as $|P_1| + |P_2| + f(P_1, P_2)$ where it can generalize to consider both penalizing common edges or nearby edges.

Has this problem been addressed before?
If the cost is $|P_1|+|P_2|+|P_1∩P_2|$, then a simple reduction to the shortest pair edge disjoint paths gives us a polynomial time solution. For each edge $e=(u,v)$ add two edges $(u,uv)$ and $(uv,v)$ each of them with same edge weight as $e$. The shortest pair edge disjoint paths in the new graph corresponds to the required solution in the original graph.

The algorithm for shortest pair edge disjoint paths is known as Suurballe's algorithm.

https://en.m.wikipedia.org/wiki/Suurballe%27s_algorithm

If the cost is some arbitrary function the problem can be NP-complete. 
e.g assuming edge weights one, let the cost function be as $|P_1| + |P_2| + |P_1∩P_2| + 2\cdot |G-P_1\cup P_2|$.
Then the cost is always $ \ge |G|+2$ (if source and terminal are distinct vertices). The equality holds iff the graph has a Hamiltonian cycle.