Step 1: Let the first term be $a$ and common difference be $d$. Then:

Third term: $a_3 = a + 2d$

Seventh term: $a_7 = a + 6d$

Step 2: Given that sum of 3rd and 7th terms is $6$:

$$(a + 2d) + (a + 6d) = 6$$ $$2a + 8d = 6$$ $$a + 4d = 3 \quad \text{... (1)}$$

Step 3: Given that their product is $8$:

$$(a + 2d)(a + 6d) = 8$$

From equation (1): $a = 3 - 4d$

Substituting:

$$(3 - 4d + 2d)(3 - 4d + 6d) = 8$$ $$(3 - 2d)(3 + 2d) = 8$$ $$9 - 4d^2 = 8$$ $$d^2 = \frac{1}{4}$$

Therefore, $d = \frac{1}{2}$ or $d = -\frac{1}{2}$

Step 4: Finding $a$ for both cases:

When $d = \frac{1}{2}$: $a = 3 - 4\left(\frac{1}{2}\right) = 1$

When $d = -\frac{1}{2}$: $a = 3 - 4\left(-\frac{1}{2}\right) = 5$

Step 5: Finding sum of first 16 terms using $S_n = \frac{n}{2}[2a + (n-1)d]$:

Case 1 ($a = 1$, $d = \frac{1}{2}$):

$$S_{16} = \frac{16}{2}\left[2(1) + 15\left(\frac{1}{2}\right)\right] = 8\left[2 + \frac{15}{2}\right] = 8 \times \frac{19}{2} = \mathbf{76}$$

Case 2 ($a = 5$, $d = -\frac{1}{2}$):

$$S_{16} = \frac{16}{2}\left[2(5) + 15\left(-\frac{1}{2}\right)\right] = 8\left[10 - \frac{15}{2}\right] = 8 \times \frac{5}{2} = \mathbf{20}$$