Let the required remaining observations be $x$ and $y$, then Mean $\bar{x}=8=\frac{\sum x_i}{7}$
$\begin{aligned}
&\Rightarrow 8=\frac{2+4+10+12+14+x+y}{7} \\
&\Rightarrow 56=42+x+y \text { or } x+y=14 \quad \ldots(i)
\end{aligned}$
i.e. Variance $=16=\frac{\sum\left(x_i-\bar{x}\right)^2}{n}$


$\Rightarrow 112=236+x^2+y^2-16(x+y)$
$\Rightarrow 112=236+x^2+y^2-16 \times 14$
$\Rightarrow x^2+y^2=100$
i.e. $(x+y)^2=x^2+y^2+2 x y \Rightarrow 14^2=100+2 x y$
i.e. $196-100=2 x y \quad \Rightarrow 2 x y=96$
So, $(x-y)^2=x^2+y^2-2 x y \Rightarrow 100-96=4$
i.e. $x-y=\pm 2 \quad \ldots(ii)$

Solving (i) and (ii), we get

are the remaining observations.