The mean of $6$ distinct observations is $6.5$ and their variance is $10.25.$ If $4$ out of $6$ observations are $2,4,5$ and $7,$ then the remaining two observations are:

$\frac{a+b+2+4+5+7}{6}=6.5$

$\Rightarrow a+b=21$

$\Rightarrow b=21-a$

Then, other two numbers are $a$ and $\left(21-a\right)$

Now, variance is $10.25$, therefore

$10.25=\frac{\left(4+16+25+49+a^2+{\left(21-a\right)}^2\right)}{6}-{\left(6.5\right)}^2$

$\Rightarrow 6\left(10.25\right)+6{\left(6.5\right)}^2=94+a^2+{\left(21-a\right)}^2$

$\Rightarrow a^2+{\left(21-a\right)}^2=221$

$\therefore a=10$ and $\left(21-a\right)=21-10=11$

So, remaining two observations are $10,11$.