The given sets are $A = \{(x, y) \in R \times R : x^2 + y^2 = 25\}$ and $B = \{(x, y) \in R \times R : x^2 + 9y^2 = 144\}$. To find the intersection of A and B, we solve the system of equations. Subtracting the equation for A from the equation for B gives: $x^2 + 9y^2 - (x^2 + y^2) = 144 - 25$, which simplifies to $8y^2 = 119$, so $y^2 = \frac{119}{8}$. Substituting this into the equation for A, we get $x^2 = 25 - \frac{119}{8} = \frac{200 - 119}{8} = \frac{81}{8}$. Thus, $x = \pm \frac{9}{\sqrt{8}}$ and $y = \pm \frac{\sqrt{119}}{\sqrt{8}}$. Since $D = A \cap B$, the set D contains 4 points. Now, consider the set $C = \{(x, y) \in Z \times Z : x^2 + y^2 \le 4\}$. The integer pairs (x, y) that satisfy this condition are: $(\pm 2, 0)$, $(0, \pm 2)$, $(\pm 1, 0)$, $(0, \pm 1)$, $(\pm 1, \pm 1)$, and $(0, 0)$. Counting these points, we find that there are 13 such pairs. The number of one-to-one functions from D to C is the number of ways to choose 4 elements from C and arrange them, which is $P(13, 4) = \frac{13!}{(13-4)!} = \frac{13!}{9!} = 13 \times 12 \times 11 \times 10 = 17160$.