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\limits_{i = 1}^8 {{{\left( {{x_i} – \bar x} \right)}^2}} $

$9.25=\frac{1}{8}[(-3)^{2}+(-2)^{2}+(1)^{2}+(3)^{2}+(4)^{2}$

$+x^{2}+y^{2}-2 \times 9(x+y)+2 \times(9)^{2}]$

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

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

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

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

From $(1),$ we obtain

$x^{2}+y^{2}+2 x y=144$        ……..$(3)$

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

$2 x y=64$      ……….$(4)$

Subtracting $(4)$ from $(2),$ we obtain

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

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

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 $8$