I presume you mean, as wilmer suggested, ((2^n)(4^(n+1))/8^(n-2). You need to know that 4= 2^2 and 8= 2^3 so 4^(n+1)= (2^2)^(n+1)= 2^(2(n+1))= 2^(2n+2) and 8^(n-2)= (2^3)^(n-2)= 2^(3(n-2))= 2^(3n- 6).
So your problem is ((2^n)(2^(2n+2))/(2^(3n- 6)). Combine those. How many "2"s do you have in the numerator and how many in the denominator?
You know that (√3+ 1)^2= (√3)^2+ 2(√3)+ 1, don't you? And (√3)^2= 3 so x^2= 2(√3)+ 4. For 1/x^2= 1/(2(√3)+ 4), "rationalize the denominator": multiply both numerator and denominator by 2(√3)- 4.Also, if x = (√3) + 1 what is the value of x^2 - 1/x^2 with the answer as a simplified surd?
(This is the third problem you have posted not showing any attempt to do them yourself. I you honestly have no idea how to even begin any of the exercises you are being given, perhaps this course is too difficult for you.)