Answer
(a) $\lambda =9.5 \times 10^{-8} \mathrm{C/m}$
(b) $\Delta V = 429 \mathrm{V}$
(c) $\Delta V = 0$
Work Step by Step
(a) The linear charge $\lambda$ could be caclauted from the potential difference in the next step \begin{gather*} \Delta V =\frac{\lambda}{2 \pi \epsilon_{o}} \ln \frac{r_{b}}{r_{a}}\\ \lambda = 2 \pi \epsilon_{o} \Delta V / \ln(r_b/r_a)\\ \lambda = \frac{575 \times 2 \pi \times 8.85 \times 10^{-12}}{\ln (3.5 / 2.5)}\\ \boxed{\lambda =9.5 \times 10^{-8} \mathrm{C/m}} \end{gather*}
(b) Now for $r_a$ = 3.5 cm and $r_b$ = 4.5 cm the potential will be \begin{align} \Delta V &= \frac{\lambda}{2 \pi \epsilon_{o}} \ln \frac{r_{b}}{r_{a}}\\ &= \frac{9.5 \times 10^{-8}}{2 \pi \times 8.85 \times 10^{-12}} \ln \left(\frac{4.5}{3.5}\right)\\ &= \boxed{429 \mathrm{V}} \end{align}
(c) The potential will be zero. $$\Delta V = 0$$
