In order to simplify this, you need to make sure #i# isn't in the denominator (or bottom) of the fraction. To do this, multiply both the numerator and denominator by the conjugate of #3-2i#, which is #3+2i#. When doing this, make sure to use the FOIL method:
1)
#((8+6i) * (3+2i))/((3-2i) * (3+2i))#
2)
#(24+16i+18i-12)/(9+6i-6i-4i^2)#
Since #i# equals #sqrt(-1)#, #i^2# equals #-1#.
3)
#(12+34i)/(9+4)#
4)
#(12+34i)/(13)#