The mean and variance of 7 observations are 8 and 16,

Let the remaining two observations be x and y.

The observations are 2, 4, 10, 12, 14, x, y.

Mean, $\bar{x}=\frac{2+4+10+12+14+x+y}{7}=8$

$\Rightarrow 56=42+x+y$

      $\Rightarrow x+y=14$ ...(1)

Variance $=16=\frac{1}{n} \sum_{i=1}^{7}\left(x_{i}-\bar{x}\right)^{2}$

$16=\frac{1}{7}\left[(-6)^{2}+(-4)^{2}+(2)^{2}+(4)^{2}+(6)^{2}+x^{2}+y^{2}-2 \times 8(x+y)+2 \times(8)^{2}\right]$

$16=\frac{1}{7}\left[36+16+4+16+36+x^{2}+y^{2}-16(14)+2(64)\right]$ [Using (1)]

$16=\frac{1}{7}\left[108+x^{2}+y^{2}-224+128\right]$

$\Rightarrow x^{2}+y^{2}=112-12=100$

$x^{2}+y^{2}=100$ $\ldots(2)$

From (1), we obtain

$x^{2}+y^{2}+2 x y=196 \ldots(3)$

$2 x y=196-100$

$\Rightarrow 2 x y=96 \ldots$ (4)

Subtracting (4) from (2), we obtain

$x^{2}+y^{2}-2 x y=100-96$

$\Rightarrow(x-y)^{2}=4$

$\Rightarrow x-y=\pm 2 \ldots$ (5)

Therefore, from (1) and (5), we obtain

x = 8 and y = 6 when x – y = 2

x = 6 and y = 8 when x – y = – 2

Thus, the remaining observations are 6 and 8.