I would like to know how much energy would it require to lift a 40-ton
weight 2.75 ft a second. I also would like to know that by using gears
or pulleys what would be the list effort we can manage the lift and
how many horsepower will require to do the job? I know that there are
many motors that can do the job. How much fuels will a diesel engine
consume lifting this weight at that rate over 75 feet? Will hydraulic
motor be a better choice for this kind of job? What would be the best
choice to do this kind of work having fuel consumptions in mind?

You need to distinguish the terms "energy" & "power"
Total energy E = Fd = 40ton * 2000 lb/ton * 75 ft = 6000000 ft-lb
Power = E/t = F*d/t = F*v = 80000 lb * 2.75 ft/sec = 220000 ft-lb/sec
In terms of horsepower 1 hp = 550 ft-lb/sec so power required 
         = 220000 ft-lb/sec / 550 ft-lb/sec = 400 hp
Now we said that that the total energy required is 6000000 ft-lb
Also 1 BTU = 778 ft-lb so the total energy in BTU 
                 = 6000000 ft-lb / 778 ft-lb/BTU = 7712 BTU
The Handbook of Chemistry and Physics says that 1 lb of common fuels
such as gasoline yield "approximately" 20000 BTU / lb so
at 100 % efficiency about 1/3 lb (7712 BTU / 20000 BTU/lb = .39 lb)
of gasoline would be required to lift the 40 tons to a height of 75 ft
Since gasoline engines operate at probably the order of 25% efficiency
.39 lb gasoline / .25 = 1.56 lbs of gasoline
Diesel engines are somewhat more efficient and would maybe only require
a little more than pound of fuel oil.
Obviously there are approximations involved here regarding
the engine efficiency and the heat of combustion of the fuel used.
Also, mechanical advantage obtained using gears, pulleys, etc. have
nothing to do with the total energy required other than that the frictional
energy lost in using such a device would require additional energy input
to overcome the energy lost thru friction.