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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 @Pergrine Rook. Thanks]
 So we have $A(1) = 3/2 ,\ A(3) = 3/4 ,\ A(5) = 3/8,\ A(7)= 3/16\cdots $
and $A(2) =2/3 , \ A(4) =1/3, \ A(6) =1/6, \ A(8)=1/12\cdots $

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}

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. $$$$ $2/3$ is irreducible, so it is the second term. $4/3 = 1 + 1/3$. Thus, $1/3$ is the fourth term.  Similarly, $8/3 = 2 + 2/3$. So, the sixth term should be $2/3$  $$$$ For answer 2: Divide the first and second terms, repeatedly by 2 $$$$ So, $3/2,3/4,3/8,3/16\cdots$ - Odd terms $$$$and $2/3,1/3,1/6,1/12,\cdots$ - Even terms

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 @Pergrine Rook. Thanks]
So we have $A(1) = 3/2 ,\ A(3) = 3/4 ,\ A(5) = 3/8,\ A(7)= 3/16\cdots $
and $A(2) =2/3 , \ A(4) =1/3, \ A(6) =1/6, \ A(8)=1/12\cdots $

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. $$$$ $2/3$ is irreducible, so it is the second term. $4/3 = 1 + 1/3$. Thus, $1/3$ is the fourth term. Similarly, $8/3 = 2 + 2/3$. So, the sixth term should be $2/3$ $$$$ For answer 2: Divide the first and second terms, repeatedly by 2 $$$$ So, $3/2,3/4,3/8,3/16\cdots$ - Odd terms $$$$and $2/3,1/3,1/6,1/12,\cdots$ - Even terms

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. $$$$ $2/3$ is irreducible, so it is the second term. $4/3 = 1 + 1/3$. Thus, $1/3$ is the fourth term. Similarly, $8/3 = 2 + 2/3$. So, the sixth term should be $2/3$ $$$$ For answer 2: Divide the first and second terms, repeatedly by 2 $$$$ So, $3/2,3/4,3/8,3/16\cdots$ - Odd terms $$$$and $2/3,1/3,1/6,1/12,\cdots$ - Even terms