Find the values of a and b if

$\frac{7+3 \sqrt{5}}{3+\sqrt{5}}-\frac{7-3 \sqrt{5}}{3-\sqrt{5}}$

$=\frac{7+3 \sqrt{5}}{3+\sqrt{5}} \times \frac{3-\sqrt{5}}{3-\sqrt{5}}-\frac{7-3 \sqrt{5}}{3-\sqrt{5}} \times \frac{3+\sqrt{5}}{3+\sqrt{5}}$

$=\frac{7(3-\sqrt{5})+3 \sqrt{5}(3-\sqrt{5})}{3^{2}-(\sqrt{5})^{2}}-\frac{7(3+\sqrt{5})-3 \sqrt{5}(3+\sqrt{5})}{3^{2}-(\sqrt{5})^{2}}$

$=\frac{21-7 \sqrt{5}+9 \sqrt{5}-15}{9-5}-\frac{21+7 \sqrt{5}-9 \sqrt{5}-15}{9-5}$

$=\frac{6+2 \sqrt{5}}{4}-\frac{6-2 \sqrt{5}}{4}$

$=\frac{6+2 \sqrt{5}-6+2 \sqrt{5}}{4}$

$=\frac{4 \sqrt{5}}{4}$

$=\sqrt{5}$

$\therefore \frac{7+3 \sqrt{5}}{3+\sqrt{5}}-\frac{7-3 \sqrt{5}}{3-\sqrt{5}}=0+1 \times \sqrt{5}$

Comparing with the given expression, we get

a = 0 and b = 1

Thus, the values of a and b are 0 and 1, respectively.