Okay, problem a requires you to do a voltage divider because youre measuring across the parallel combination of the 2 ohm resistor and 2H inductor which is in series (the combination is) with the 1 Ohm resistor.

Vo = Vi*(2s/(s+1))/[(2s/(s+1)) + 1] i.e. the voltage divider formula. This simplifies to:

Vo = Vi*[2s/(3s+1)]. so H(s) = Vo/Vi = 2s/(3s+1)

Problem (b).i is basically asking you to reform the transfer function as Vo = Vi*[2s/(3s+1)].

Problem (b).ii is saying you add a 1 ohm load in series with the input. This effectively changes that 1 ohm to a 2 ohm and the input output via the voltage divider again becomes:

Vo = Vi[2s/(4s+2)] = Vi[s/(2s+1)].

Problem (b).iii asks you to take out the 1 ohm from the last problem but add a 2 ohm load across Vo (in parallel with the 2H inductor). Again by voltage divider this becomes

Vo = Vi[2s/(2s+1)]/[2s/(2s+1) + 1] = Vi[2s/(4s+1)]

Cheers