Path finding under constraints
One solution would be to use integer linear programming (ILP).  This is not guaranteed to have a polynomial worst-case running time (ILP can take exponential time), but ILP solvers are pretty impressive, and for reasonable-sized graphs, this might just be an effective solution in practice.

For each edge $e \in E$, let $x_e,y_e,z_e$ be integer variables with the constraint $0 \le x_e \le 1$, $0 \le y_e \le 1$, $0 \le z_e \le 1$ (so that $x_e,y_ez_e$ are forced to be either 0 or 1).  The meaning of $x_e=1$ is that $e$ is part of the path $l_1$, and the meaning of $y_e=1$ is that $e$ is part of $l_2$, and the meaning of $z_e$ is that $e$ is part of both $l_1$ and $l_2$.

Now introduce the following constraints:


$\sum_{(a,v) \in E} x_{(a,v)} = 1$ (i.e., at least one of the edges out of $a$ must be selected in $l_1$).
$\sum_{(v,b) \in E} x_{(v,b)} = 1$ (i.e., $l_1$ contains an edge into $b$).
For each vertex $v \in V \setminus \{a,b\}$, $\sum_{(u,v) \in E} x_{(u,v)} = \sum_{(v,w) \in E} x_{(v,w)}$ (if there's an edge entering $v$, there has to be an edge leaving $v$).
$\sum_{(b,v) \in E} y_{(b,v)} = 1$ (i.e., $l_2$ contains an edge out of $b$).
$\sum_{(v,a) \in E} y_{(v,a)} = 1$ (i.e., $l_1$ contains an edge into $a$).
For each vertex $v \in V \setminus \{a,b\}$, $\sum_{(u,v) \in E} y_{(u,v)} = \sum_{(v,w) \in E} y_{(v,w)}$ (if there's an edge entering $v$, there has to be an edge leaving $v$).
$\sum_{e \in E} w(e) (x_e + y_e) \le 1.1 (\alpha+\beta)$ (condition 3 in the question holds).
For each $e \in E$, $z_e \ge x_e + y_{-e} - 1$, $z_e \le x_e$, and $z_e \le y_{-e}$, where $-e$ is the reversal of the edge $e$ (i.e., if $e=(u,v)$, then $-e=(v,u)$).  (This enforces that $z_e = 1$ if and only if $x_e = y_{-e} = 1$.)


Now minimize $\sum_{e \in E} w(e) z_e$.  This will minimize the total weight of the edges that are common to both $l_1$ and $l_2$.