(a) Tâ‚‚ = 126Â°C , Tâ‚‚s = 106Â°C(b) 64%Explanation:we will begin with a step by step process;using the expression WÑ‚ = m(ex)Cp(Tâ‚ - Tâ‚‚)where WÑ‚ = turbine power outputm(ex) = mass of the exhaust gasCp = specific heat at constant pressure Tâ‚ = exhaust gas temperature at the turbine inletTâ‚‚ = exhaust gas temperature at the turbine exitusing from the gas table, the specific heat of exhaust gas at avg temperature of 425Â°C as 1.075kJ/kg.Kmaking substitutions we have;Tâ‚ = 450Â°C, Tâ‚‚ = 400Â°C, m(ex) = 0.02kg/s, Cp = 1.075kJ/kg.KSubstituting values into formula gives;WÑ‚ = m(ex)Cp(Tâ‚ - Tâ‚‚) WÑ‚ = [(0.02kg/s)(1.075kJ/kg.K)(450+273)K-(400+273)K]WÑ‚ = 1.075 kWThe relationship of mechanical efficiency between turbine and compressor is given as;Ð›Ð¼ = Wc/WÑ‚ (1)where WÑ‚ = turbine output Â Â Â Â Â Â Wc = compressor output Â Â Â Â Â Â Ð›Ð¼ = mechanical efficiency Wc = Ð›Ð¼ Ã— WÑ‚ Wc = 0.95 Ã— 1.075(kW)Wc = 1.021 kWcompressor power:Wc = m(air)Cp(Tâ‚‚ - Tâ‚)where m(air) = mass of airTâ‚ = air temp at the compressor inlet Tâ‚‚ = air temp at the compressor exit Cp = specific heat at constant pressure Tâ‚‚ = Tâ‚ + Wc/m(air)Cp (2)the specific heat of air at the expected average temperature of 100Â°C is 1.011 kJ/kg.Ksubstituting Tâ‚ = 70Â°C, Wc = 1.021 kW, m(air) = 0.018 kg/sm Â Cp = 1.011 kJ/kg.KTâ‚‚ = 70 Â°C + (1.021)kW/ (0.018kg/s)(1.011 kJ/kg.K)Tâ‚‚ = 126 Â°C Considering an isentropic process, the air temperature at the compressor exit; Tâ‚‚s = Tâ‚ (PÂ²/PÂ¹)âˆ§(k-1)/(k)where Tâ‚ = 70Â°C, Pâ‚= 95kPa, Pâ‚‚ = 135kPa, k = 1.397Tâ‚‚s = (70+273)K (135/95 kPa) âˆ§ (1.397-1)/(1.397)Tâ‚‚s = 379 K = 106 Â°C(b). the isentropic efficiency of the compressor Â Â Â Ð›c = (Tâ‚‚s - Tâ‚ / Tâ‚‚ - Tâ‚) Â Â Â Â Ð›c = (106Â°C - 70Â°C) / (126Â°C - 70Â°C) Â Â Â Â Â Â Ð›c = 0.642 = 64%the isentropic efficiency of the compressor is 64%cheers i hope this helps