I don't know, what you are referring to here. The problems mentioned in the book are all of the form that you have a given spherically symmetric charge distribution ##\rho(r)## and want the electrostatic field everywhere. For that it's most easy to calculate the electrostatic potential, which by symmetry is a function of ##r## only (in spherical coordinates), i.e.,
$$-\frac{1}{r} [r \Phi(r)]''=\frac{1}{\epsilon_0} \rho(r).$$
The charge distribution is is assumed to vanish outside a sphere of radius ##R##, i.e., ##\rho(r)=0## for ##r>R##.
So for ##r>R## you get
$$[r \Phi(r)]''=0 \; \Rightarrow \; r \Phi(r)=C_1+C_2 r \; \Rightarrow \; \Phi(r)=\frac{C_1}{r} + C_2 \quad \text{for} \quad r>R.$$
We can choose an arbitrary additive constant anyway freely, and it's convenient to choose it such that ##\Phi(r) \rightarrow 0## for ##r \rightarrow \infty##, i.e., we set ##C_2=0##. For the electric field you get there
$$\vec{E}=-\vec{\nabla} \Phi=\frac{C_1}{r^2} \vec{e}_r.$$
In the charge-free space you have a Coulomb field, and Gauß's Law tells you that for any spherical shell with a radius ##a>R## you have
$$\int_{S(a)} \mathrm{d}^2 \vec{f} \cdot \vec{E}=4 \pi a^2 E_r(a)=4 \pi C_1=\frac{Q}{\epsilon_0} \; \Rightarrow \ \; C_1=\frac{Q}{4 \pi\epsilon_0}.$$
For ##r<R## you have to integrate the above equation including the given ##\rho(r)## twice, i.e.,
$$[r \Phi(r)]'=-\int_0^r \mathrm{d} r' r' \rho(r')/\epsilon_0 + C_1',$$
and then
$$r \Phi(r)=-\int_0^r \mathrm{d} r'' \int_0^{r''} \mathrm{d} r ' r' \rho(r')/\epsilon_0 + C_1' r + C_2'$$
or
$$\Phi(r)=-\frac{1}{r} \int_0^r \mathrm{d} r'' \int_0^{r''} \mathrm{d} r ' r' \rho(r')/\epsilon_0 + C_1'+\frac{C_2'}{r} \quad \text{for} \quad r<R.$$
Now there are no singularities at ##r=0##, i.e., you must have ##C_2'=0##. Further there's no surface charge at ##r=R## in addition to the "bulk" charge distribution ##\rho(r)## for ##r<R##, and thus both ##E_r## and ##\Phi## are continuous at ##r=R##, from which you get ##C_1'##. For ##\vec{E}=E_r \vec{e}_r## you even don't need ##C_1'## at all.
Now take the first example,
$$\rho(r)=\rho_0=\frac{3 Q}{4 \pi R^3} \quad \text{f"ur} \quad r<R.$$
Then for ##r<R## you have
$$[r \Phi(r)]''=-\frac{\rho_0}{\epsilon_0} r \; \Rightarrow \; [r \Phi(r)]'=-\frac{\rho_0}{2 \epsilon_0} r^2 + C_1' \; \Rightarrow \; r \Phi(r)=-\frac{\rho_0}{6 \epsilon_0} r^3 + C_1' r.$$
Finally
$$\Phi(r)=-\frac{\rho_0}{6 \epsilon_0} r^2 + C_1'=-\frac{Q}{8 \pi \epsilon_0 R^3} r^2+C_1' \quad \text{for} \quad r<R.$$
For the electric field you thus get
$$E_r=-\Phi'(r)=\begin{cases} \frac{Q}{4 \pi \epsilon_0 R^3} r & \text{for} \quad r<R, \\ \frac{Q}{4 \pi \epsilon_0 r^2} & \text{for} \quad r \geq R, \end{cases}$$
and indeed ##E_r## is continuous at ##r=R##, as already argued about above.
The 2nd example, where ##\rho=C r## for ##r<R## you can try to solve for yourself. It's a good exercise!
