A galvanometer has a coil resistance of 60 Ω and a full-scale current of 5.0 mA…

Step 1. The voltage. At full scale the coil has $V = I_gG = 5.0 \times 10^{-3}\ \text{A} \times 60\ \Omega = 0.30\ \text{V}$.

Step 2. The $0.50$ A range. The shunt carries $0.50\ \text{A} - 0.005\ \text{A} = 0.495\ \text{A}$. So $S = \dfrac{0.30\ \text{V}}{0.495\ \text{A}} = 0.606\ \Omega$.

Step 3. The $3.0$ A range. The shunt carries $3.0\ \text{A} - 0.005\ \text{A} = 2.995\ \text{A}$. So $S = \dfrac{0.30\ \text{V}}{2.995\ \text{A}} = 0.100\ \Omega$.

Step 4. The resistance of the ammeter. $R_A = \dfrac{V}{I} = \dfrac{0.30\ \text{V}}{3.0\ \text{A}} = 0.100\ \Omega$. It is a little less than the shunt, because the coil is in parallel with it.