Let's first be clear about some basics then we go on to questions:
The periodic boundary condition requires that any wave in the sample $e^{ikr}$ have the same value for a position $r$ as it has for $r+L$ (after one round around the circle). This imposes quantization of $k$
$$k=\frac{2\pi n}{L} \ \ \ n\ \text{is integer}.$$
In three dimensions,
$$\mathbf{k}=\frac{2\pi }{L}(n_1,n_2,n_3)$$
The above picture taken from Blundell shows what I said in a two-dimensional case.
So here each $\mathbf{k}$ point now occupies a volume of $(2\pi /L)^3$. Because of this discretization of values of $\mathbf{k}$, whenever we have to sum over all possible $\mathbf{k}$ we obtain
$$\sum_\mathbf{k}\rightarrow \frac{L^3}{(2\pi)^3}\int \mathbf{dk}$$
At $T=0$ for electrons,
$$N=2\times\frac{L^3}{(2\pi)^3} \int \mathbf{dk} \theta(E_F-\epsilon(\mathbf{k}))=2\times\frac{L^3}{(2\pi)^3} \int^{|k|<k_F} \mathbf{dk} =2\frac{L^3}{(2\pi)^3}\left(\frac{4}{3}\pi k_F^3\right)$$
where the prefactor of $2$ accounts for the two possible spin states each possible wavevector $\mathbf{k}$.
Problem ($1$)
As stated earlier,
$$\mathbf{k}=\frac{2\pi }{L}(n_1,n_2,n_3)$$
As you can see from the figure as we go up the density of states increases which allows us to use integrals.
What you are stating in question is wrong.
Problem ($2$)
This has been answered above. The word orbitals are not right in this situation we are dealing with free electron gas. What we have found is the number of electrons not the number of orbital or something.
Problem ($3$)
The free electron gas is not bound to atoms so we are not in concert with a magnetic quantum number. There can be two spin-state correspondings to a partial wave function.
Edit: If we focus on the magnitude of the wave vector given by $k$. Allowed states with a wave vector whose magnitude lies between $k$ and $k+dk$ given by the density of states:
$$g(k)dk=\frac{\text{volume in k-space of a spherical shell}}{\text{volume in k-space occupied per allowed state}}$$
$$g(k)dk=\frac{4\pi k^2dk}{(2\pi /L)^3}\propto k^2dk$$
That's what I meant by increasing the density of state.