(1)[(1/4)^4005]/(16^2003)=[1/(4^4005)]*(4^2)^2003=(4^4006)/(4^4005)=4
(2)8(5x+3/2)[(5/2)x)-3/4]=16[(5/2)x)+3/4][(5/2)x-3/4]
=16[(25/4)x^2-9/16]=100x^2-9