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.