The mean and variance of 8 observations are

Let the two remaining observations be $x$ and $y$.

$\because \bar{x}=\frac{5+7+10+12+14+15+x+y}{8}$

$\Rightarrow 10=\frac{63+x+y}{8}$

$\Rightarrow x+y=80-63$

$\Rightarrow x+y=17$ ...(i)

$\because \operatorname{var}(x)=13.5$

$=\frac{25+49+100+144+196+225+x^{2}+y^{2}}{8}-(10)^{2}$

$\Rightarrow x^{2}+y^{2}=169$.....(ii)

From (i) and (ii) we get

$(x, y)=(12,5)$ or $(5,12)$

So, $|x-y|=7$.