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:

Let the two unknown observations be $$a$$ and $$b$$. The six observations are $$2, 4, 5, 7, a, b$$.

From the mean: $$\frac{2+4+5+7+a+b}{6} = 6.5$$, so $$18 + a + b = 39$$, giving $$a + b = 21$$.

From the variance formula $$\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$$: $$10.25 = \frac{\sum x_i^2}{6} - (6.5)^2 \implies \frac{\sum x_i^2}{6} = 10.25 + 42.25 = 52.5 \implies \sum x_i^2 = 315.$$

The sum of squares of the known observations is $$4 + 16 + 25 + 49 = 94$$, so $$a^2 + b^2 = 315 - 94 = 221$$.

Now using $$(a+b)^2 = a^2 + 2ab + b^2$$: $$441 = 221 + 2ab$$, so $$ab = 110$$.

The values $$a$$ and $$b$$ are roots of $$t^2 - 21t + 110 = 0$$, which factors as $$(t-10)(t-11) = 0$$.

Therefore the two remaining observations are $$10$$ and $$11$$.