Ages

$\frac{A}{B} = \frac{8}{11}$ Then, $A = \frac{8}{11}B$
$\frac{A+8}{B+8} = \frac{4}{5}$ Then, $5A+40 = 4B+32$
$5A+8 = 4B$ Substitute A as $\frac{8}{11}B$
Solving this equation give us B=22, A=16. Thus, the difference between the ages is $\boxed{6}$

$\dfrac{A}{B}=\dfrac{8}{11}$ $\implies$ $A=\dfrac{8}{11}B$ $\color{#D61F06}(1)$

Eight years later, the ratio becomes

$\dfrac{A+8}{B+8}=\dfrac{4}{5}$ $\implies$ $5A+40=4B+32$ $\color{#D61F06}(2)$

substitute $\color{#D61F06}(1)$ in $\color{#D61F06}(2)$

$5\left(\dfrac{8}{11}B\right)+40=4B+32$

$B=22$

It follows that,

$A=\dfrac{8}{11}(22)=16$

The difference of their ages is $22-16=$ $\boxed{6}$