Answer
a) $\frac{\lambda}{2\pi\epsilon_or}$
b) $\frac{\lambda}{2\pi\epsilon_or}$
c) Attaced image.
Blue: r < a
Red: a < r < b
Yellow: b < r < c
Purple: c < r < 2c
d) Inner surface= $-\lambda$ Outer surface= $+\lambda$
Work Step by Step
a) Use Gauss's law for a length "L" of the cable, $E.A=\frac{Q_e}{\epsilon_o}$ Area os the Gaussian Cylinder would be then given by: $A= 2.\pi.r.L$ And the charge enclosed in the surface would be $Q_e=\lambda.L$ Combining these three equations we get: $E.2.\pi.r.L=\frac{\lambda.L}{\epsilon_o}$ $E=\frac{\lambda.L}{\epsilon_o.2.\pi.r.L}$ $E=\frac{\lambda}{2.\epsilon_o.\pi.r}$ where [a < r < b]
b) Using the same method as in a) we get $E=\frac{\lambda}{2.\epsilon_o.\pi.r}$ where r > b
c) We have obtained the expressions for the magnitude of the field for a distance [a < r < b] and [r > b] from the center of the coaxial cable, which is basically $E=\frac{A}{r}$ where A is a constant. The value of the field inside a conductor is always zero. That would give E=0 for [r < a] and [c < r < b].
d) We know unlike charge attract each other. Since the inner cylinder is positively charged the inner walls of the outer cylinder will have the equal but negative charge density. Hence -$\lambda$. and since we know that the total charge of the larger cylinder is zero, the outer wall of the said cylinder will be equal to but the negative of the inner side. Hence -(-$\lambda$)=+$\lambda$
