The mean and variance of eight observations are 9 and 9.25, respectively.

Solution:

Let the remaining two observations be x and y.

Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y.

Mean, $\bar{x}=\frac{6+7+10+12+12+13+x+y}{8}=9$

$\Rightarrow 60+x+y=72$

     $\Rightarrow x+y=12$ ...(1)

Variance $=9.25=\frac{1}{n} \sum_{i=1}^{8}\left(x_{i}-\bar{x}\right)^{2}$

$9.25=\frac{1}{8}\left[(-3)^{2}+(-2)^{2}+(1)^{2}+(3)^{2}+(3)^{2}+(4)^{2}+x^{2}+y^{2}-2 \times 9(x+y)+2 \times(9)^{2}\right]$

$9.25=\frac{1}{8}\left[9+4+1+9+9+16+x^{2}+y^{2}-18(12)+162\right]$ [Using (1)]

$9.25=\frac{1}{8}\left[48+x^{2}+y^{2}-216+162\right]$

$9.25=\frac{1}{8}\left[x^{2}+y^{2}-6\right]$

$\Rightarrow x^{2}+y^{2}=80$ $\ldots(2)$

From (1), we obtain

$x^{2}+y^{2}+2 x y=144 \ldots$ (3)

From $(2)$ and $(3)$, we obtain

$2 x y=64 \ldots(4)$

Subtracting (4) from (2), we obtain

$x^{2}+y^{2}-2 x y=80-64=16$

$\Rightarrow x-y=\pm 4 \ldots(5)$

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

$x=8$ and $y=4$, when $x-y=4$

$x=4$ and $y=8$, when $x-y=-4$

Thus, the remaining observations are 4 and 8.