double angle formula computations $\cos A = 2/\sqrt5$ $(3\pi/2 < A < 2\pi)$ and $\sin B = 4/5$ $(\pi/2 < B < \pi)$, compute:
$A=\alpha$,
$B=\beta$
a) $\sin(A-B)$
b) $\cos(B/2)$
c) $\tan2A + \tan2B$
d) $\cos2A - \sin4B$
e) $\tan(3A)$
for a) I got $-\sqrt5/5$ with $\sin(A)= -1/\sqrt5$ and $\cos(B)= -3/5$
I do not remember how to do the rest. 
 A: Hints:
You have to use the basic trigonometric identities:
$$\sin(A\pm B)=\sin A\cos B\pm\sin B\cos A$$
$$\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$$
$$\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}$$
$$\sin A=\pm\sqrt{1-\cos^2A}\;,\;\;\text{the sign depending on the angle, for example:}$$
$$\cos A=\frac2{\sqrt5}\;\;,\;\frac{3\pi}2<A<2\pi \;\;\;\text{(and thus we're in the fourth quadrant)}\;\implies$$
$$\implies \sin A=-\sqrt{1-\frac45}=-\frac1{\sqrt5}\;,\;\;etc.$$
A: Wikipedia has a nice list of trig identities here.  Given sine or cosines of an angle, the Pythagorean identity yields the absolute value of the other function, and the sign of the function is determined by knowing which quadrant your angle lies in. This is the first step to solving the problem, and something you have already done. 
Once you know the sine and cosines of an angle, their ratio yields the tangent. Now, you will need to search for a few different identities: a sum identity, a half angle identity, a few double angle identities (one of which you will need to apply twice) and you will be able to do the first four. For the last one, use the sum identity for tangent twice, once to find $\tan(2A)$, and a second time to find $\tan(2A+A)$. 
There are some shortcuts for remembering or deriving some of the identities, but in this particular case, there are no special tricks to applying them. Beyond what you have already done , the problem is entirely looking up and applying formulae. 
