Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know, for closed surfaces, that
$$\oint EdA=\dfrac{Q_{in}}{\epsilon_0}$$
Thus, in our case,
$$EA_{sphere}=\dfrac{Q_{in}}{\epsilon_0}$$
Hence,
$$E=\dfrac{Q_{in}}{\epsilon_0A_{sphere}}=\dfrac{\Phi}{A_{sphere}}\tag 1$$
$\bullet\bullet$ When $r\lt a$,
$$\Phi_1=\dfrac{Q_{in}}{\epsilon_0} =\dfrac{Q}{\epsilon_0}$$
Plug into (1),
$$E=\dfrac{Q}{\epsilon_0(4\pi r^2)} $$
$$\boxed{\vec E_1=\dfrac{1}{ 4\pi \epsilon_0 } \dfrac{Q}{r^2}\;\hat r}$$
$\bullet\bullet$ When $b\gt r\gt a$,
$$\Phi_2=\dfrac{Q_{in}}{\epsilon_0}$$
Hence,
$$\Phi_2=\dfrac{Q+(-Q)}{\epsilon_0}=0$$
See the explanation below, and see the second figure below as well.
Thus,
$$\boxed{\vec E_2=0\;\rm N/C}$$
$\bullet\bullet$ When $r\lt a$,
$$\Phi_3=\dfrac{Q_{in}}{\epsilon_0}=\dfrac{Q+2Q}{\epsilon_0}=\dfrac{3Q}{\epsilon_0}$$
Plug into (1),
$$E_3=\dfrac{3Q}{\epsilon_0(4\pi r^2)} $$
$$\boxed{\vec E_3=\dfrac{1}{ 4\pi \epsilon_0 } \dfrac{3Q}{r^2}\;\hat r}$$
$$\color{blue}{\bf [b]}$$
Let's assume that the dashed circles in the figures below are Gaussian surfaces.
The positive charge in the cavity will pull the free electrons from inside the conductor toward the inner walls, as we see in the second figure below.
The net charge inside this Gaussian surface of $b\gt r\gt a$ is then zero since the electrons pulled toward the positive charge are creating the same charge magnitude.
This means that the inner surface of the conductor is charged by $-Q$. Note that this $-Q$ is pulled from the outer part of the hollow sphere which means that the outer walls now are charged by $+Q$ plus the original $+2Q$ charge.
Hence, the net charge of the pouter walls is now $+3Q$
$$\boxed{Q_{inner}=-Q}$$
$$\boxed{Q_{outer}=+3Q}$$
