1. Write the general terms of the two A.P.s

   • First A.P.: $3, 7, 11, \dots, 407$

     Here, $a_1 = 3, \quad d_1 = 4$ So its general term is $T_n = 3 + (n-1)4 = 4n - 1$

   • Second A.P.: $2, 9, 16, \dots, 709$

     Here, $a_2 = 2, \quad d_2 = 7$ So its general term is $S_m = 2 + (m-1)7 = 7m - 5$

2. Find the condition for common terms

   A number common to both A.P.s must satisfy $4n - 1 = 7m - 5$ Rearranging, $4n = 7m - 4$ or $4n - 7m = -4$

   We can also write the common term as a number $x$ such that $x \equiv 3 \pmod{4}$ and $x \equiv 2 \pmod{7}$

3. Solve the congruences

   Let $x = 2 + 7k$ Then we need $2 + 7k \equiv 3 \pmod{4}$ Since $7 \equiv 3 \pmod{4}$, this gives $2 + 3k \equiv 3 \pmod{4}$ $3k \equiv 1 \pmod{4}$

   Since $3^{-1} \equiv 3 \pmod{4}$, $k \equiv 3 \pmod{4}$

   So let $k = 4t + 3$ Then $x = 2 + 7(4t+3) = 2 + 28t + 21 = 23 + 28t$

   Hence, the common terms form an A.P.: $23, 51, 79, \dots$ with common difference $\operatorname{lcm}(4,7) = 28$

4. Find how many such common terms lie in the given ranges

   Since the first A.P. goes up to $407$ and the second up to $709$, the common terms must be within both lists. So the largest possible common term is $\min(407,709) = 407$

   Thus we count terms in $23, 51, 79, \dots, \le 407$

   Let the number of common terms be $N$. Then $23 + (N-1)28 \le 407$ $(N-1)28 \le 384$ $N-1 \le \frac{384}{28} = 13.714\dots$

   So, $N-1 = 13$ $N = 14$

5. Check last term

   The 14th term is $23 + 13\cdot 28 = 23 + 364 = 387$ The 15th term would be $23 + 14\cdot 28 = 415 > 407$ So exactly $14$ common terms exist.

Final Answer

$\boxed{14}$