The mean and variance of 7 observations are 8 and 16 respectively. If two observations are 6 and 8, then the variance of the remaining 5 observations is:

We have 7 observations whose mean is given as $$\bar{x}=8$$. The definition of the mean says

$$\bar{x}=\dfrac{\sum_{i=1}^{7}x_i}{7}\;.$$

Substituting $$\bar{x}=8$$, we obtain

$$8=\dfrac{\sum_{i=1}^{7}x_i}{7}\;.$$

Multiplying both sides by 7,

$$\sum_{i=1}^{7}x_i = 7 \times 8 = 56\;.$$

Out of these 7 observations, two are specifically known: $$6$$ and $$8$$. Removing them from the total, the sum of the remaining 5 observations is

$$\sum_{i=1}^{5}y_i = 56-6-8 = 42\;.$$

Now we need their variance. First we recall the population variance formula (division by the number of observations):

$$\sigma^{2}=\dfrac{\sum x_i^{2}}{n}-\bar{x}^{2}\;.$$

The variance of all 7 observations is given as $$16$$, so with $$n=7$$ we write

$$16=\dfrac{\sum_{i=1}^{7}x_i^{2}}{7}-8^{2}\;.$$

Adding $$8^{2}=64$$ to both sides,

$$16+64=\dfrac{\sum_{i=1}^{7}x_i^{2}}{7}\;,$$

$$80=\dfrac{\sum_{i=1}^{7}x_i^{2}}{7}\;.$$

Multiplying by 7,

$$\sum_{i=1}^{7}x_i^{2}=7 \times 80 = 560\;.$$

Next, we subtract the squares of the two known observations to isolate the sum of squares of the remaining five. Their squares are $$6^{2}=36$$ and $$8^{2}=64$$, whose sum is $$36+64=100$$. Therefore,

$$\sum_{i=1}^{5}y_i^{2}=560-100=460\;.$$

For the five residual observations we already found that their sum is $$42$$, so their mean is

$$\bar{y}=\dfrac{42}{5}=8.4\;.$$

Using the same variance formula for these five observations (now $$n=5$$), we have

$$\sigma_{y}^{2}=\dfrac{\sum_{i=1}^{5}y_i^{2}}{5}-\bar{y}^{2}\;.$$

Substituting the values,

$$\sigma_{y}^{2}=\dfrac{460}{5}-(8.4)^{2}\;.$$

Computing each term:

$$\dfrac{460}{5}=92,\qquad (8.4)^{2}=70.56\;.$$

Hence,

$$\sigma_{y}^{2}=92-70.56=21.44\;.$$

Now we change $$21.44$$ to an exact fraction. Observe that

$$21.44 = 21 + 0.44 = 21 + \dfrac{44}{100} = 21 + \dfrac{11}{25}= \dfrac{525}{25}+\dfrac{11}{25}= \dfrac{536}{25}\;.$$

Thus the variance of the remaining five observations is

$$\boxed{\dfrac{536}{25}}\;.$$

Hence, the correct answer is Option D.