Answer
a. $I=0.750I_0$
b. 80 nm.
Work Step by Step
a. Use $I=I_0cos^2(\phi/2)$
$$I=I_0cos^2(30.0^{\circ})=0.750I_0$$
b. Use $\phi=(\frac{2\pi}{\lambda})(r_2-r_1)$ to relate the phase to the path difference.
$$60.0^{\circ}=\frac{\pi}{3}rad$$
$$\frac{\pi}{3}rad =(\frac{2\pi}{480nm})(r_2-r_1)$$
$$r_2-r_1=80nm$$
