Sorry Darren, but this is way out of my depth.
I do normal mathematics. Wish I could help you out.
Just hang in there; someone will be along to help you out 
l) okay substitute the value of 0 in the function-->
therefore f(0)=e^-1=0.368
therefore f(x)>0.368
we substitute the value of 0 because this is the least value in the domain, i know that 0 is not included, you can take something like 0.000000000000000001, but 0 is better. or we can use differentiation to find the minimum value of the function, d/dx (e^(x-1))=e^(x-1), therefore at 0 it is 0.368, therefore since this is the minimum value every value of y will be greater than 0.368, therefore the range is f(x)>0.368
ii) let f(x)=y,
now y=e^(x-1)
take log on both sides (to find the inverse we have to make x the subject)
ln y=x-1
x=ln y + 1
therefore f^-1(x)=ln x + 1
iii) the domain of f^-1(x)=range of f(x), therefore the domain is x>0.368
