Since #a \in (\pi,(3\pi)/2)#, the angle #a# is in Quadrant III, where the #x# (length of the adjacent side) and #y# (length of the opposite side) are both negative. Denote #z# as the length of the hypotenuse. (#z# is always positive.)
We have that #sin a=-4/5 \Leftrightarrow y/z=(-4)/5#.
Note that the ratio of #y# to #z# is #-4# to #5#, but we don't necessarily have #y=-4# and #z=5#. However, we can assume so because we work with ratios (#sin, cos, tan# of angles) and not exact lengths of a triangle.
By the Pythagorean theorem,
#x^2+y^2=z^2#
#\Rightarrow x^2+(-4)^2=5^2#
#\Rightarrow x^2+16=25#
#\Rightarrow x^2=9#
#\Rightarrow x=-3# (refer to first line, #x# must be negative in Quadrant III)
So #tan a=y/x=(-4)/(-3)=4/3#.
Now use the double-angle identity #tan(2x)=(2tanx)/(1-tan^2x)#.
For #x=a/2#, the identity becomes
#tan a=(2tan(a/2))/(1-tan^2(a/2)#
#\Rightarrow 4/3=(2tan(a/2))/(1-tan^2(a/2)#
#\Rightarrow 2/3=(tan(a/2))/(1-tan^2(a/2)#
#\Rightarrow 2-2tan^2(a/2)=3tan(a/2)#
#\Rightarrow 0=2tan^2(a/2)+3tan(a/2)-2#
#\Rightarrow 0=(2tan(a/2)-1)(tan(a/2)+2)#
#\Rightarrow (2tan(a/2)-1)=0# or #tan(a/2)+2=0#
#\Rightarrow tan(a/2)=1/2# or #tan(a/2)=-2#.
