To solve this problem, think about what changes.
The air pressure decreases, so the force/pressure on the liquid columns changes. The pressure in the brine pipe does not change, and the pressure due to the water between the air and the mercury does not change. The levels of mercury do change, but the volume of mercury does not (one can assume it is incompressible).
The static head due to the mercury changes, because the elevations of the interfaces change.
Since the volume of mercury does not change, the change in mercury volume in pipe A2 must equal the change in volume in the tank A1, so one has
V2 = V1 or A2h2 = A1h1, so one has a relationship between h2 and h1 in terms of the areas.
Now, one needs a second equation, which comes from the head, H.
H = h2 + h1 for the mercury.
The static head is simply the pressure due to a height, z, of liquid and is given by [itex]\rho[/itex]gz, where [itex]\rho[/itex] is the fluid density (which is related to s.g.) and g is the acceleration of gravity.
Also think about the water (s.g. =1)/brine (s.g.=1.1) displacement as well. What is the effect of that compared to the mercury.