Answer:
a) 1 
b) 1813.96 MJ/kmol
c) 32.43 MJ/kg ,  1980.39 MJ/Kmol
Explanation:
molar mass of  ethanol (C2H5OH) = 46 g/mol
molar mass of   octane (C8H18) = 114 g/mol
therefore the moles of ethanol and octane 
ethanol =  0.85 / 46 
octane = 0.15 / 114 
a) determine the molar air-fuel ratio and air-fuel ratio by mass
attached below 
mass of air / mass of fuel = 12.17 / 1 = 12.17 
b ) Determine the lower heating value 
LHV  of  ( C2H5OH) = 26.8 * 46 = 1232.8 MJ/kmol
LHV  of (C8H18). = 44.8 mj/kg * 114 kg/kmol = 5107.2 MJ/Kmol
LHV ( MJ/kmol)  for fuel mixture = 0.85 * 1232.8 + 0.15 * 5107.2 = 1813.96 MJ/kmol
c) Determine higher heating value  ( HHV )
HHV of (C2H5OH) = 29.7 * 46 = 1366.2 MJ/kmol
HHV of C8H18 = 47.9 MJ/kg * 114 = 5460.6 MJ/kmol 
HHV  in MJ/kg  = 0.85 * 29.7 + 0.15 * 47.9  = 32.43 MJ/kg 
HHV in  MJ /kmol  =  0.85 * 1366.2 + 0.15 * 5460.8 = 1980.39 MJ/Kmol