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#.