I think the answer is
$\frac{2}{3}$ or $\frac{1}{6}$
Reason
For answer 1:
Each even term is the reciprocal of the preceding odd term. If the fraction is still reducible, then the remainder is the even term.
$2/3$ is irreducible, so it is the second term. $4/3 = 1 + 1/3$. Thus, $1/3$ is the fourth term.
Similarly, $8/3 = 2 + 2/3$. So, the sixth term should be $2/3$ $$$$
For answer 2:
$A_n = \frac{1}{2}A_{n-2}$ [a simpler version posted by @Peregrine Rook. Thanks]
So we have \begin{array}lA_1 = 3/2\\A_2 = 2/3\\A_3 = \frac12A_1 = \frac12\times 3/2=3/4\\A_4 = \frac12A_2 = \frac12\times 2/3=1/3\\A_5 = \frac12A_3 = \frac12\times 3/4=3/8\\A_6 = \frac12A_4 = \frac12\times 1/3=1/6 \text{(the answer)}\phantom{WWWWWWWWWWWWWWWW}\\ ~~\vdots\end{array}
