Given that log3 (c ) = m and log27 (d )= n , express c /(d^1/2) in the form 3^y, where y is an expression in terms of m and n.

I find that a lot of maths at A-Level is unmotivated, and this causes a lot of unnecessary confusion. You can do this question, you just need to recognise the point of log and keep a level head. Log is defined as the inverse function to the exponential, which is a lot of scary words that basically boils down to recognising that when you see log3(c)=m, and they want you to find something involving c, that you should write down c=3^m. Then similarly write d=(27)^n.
Now this looks a little scary as you have a 27 involved, what does that have to do with 3? Now you have to see that 27=3^3, and then you can see that d=3^(3n). Now c/(d^1/2)=3^(m-(1/2)*(3n))=3^(m-(3/2)n) (by properties of powers) and we're happy.