Assume a tube closed at the top (a Torricelli tube) and neglect the vapour pressure of the mercury. Now apply Pascal's Law on hydrostatic pressure:
With $P_0\approx 0$ (the pressure for $h=0$), $\rho$ the density of the mercury and $h$ the height of the measuring point, measured from the top.
The question has been made really easy by expressing the hydrostatic pressure in mm Hg:
What is the pressure 5 mm Hg below this point?
Well, this means that $h$ has been increased by 5 mm, so that the pressure at that point is simply $74+5=79$ mm Hg.
Of course this also holds if $P_0>0$. E.g. if the tube was closed at the bottom (thus forming a vessel) and open at the top, with the mercury surface exposed to some constant
pressure $P_0$, the above would also hold: the pressure would be 79 mm Hg, 5 mm below the point where it was 74 mm Hg.
We can show this as follows, with Pascal:
Pressure at the 1st point (74 mm Hg):
Pressure at the 2nd point:
Because $h_2=h_1+5$, then in mm Hg,
$P_2=P_1+5=$79 mm Hg.