When does the least squares solution exist?
$$\mathbf{A}x = b$$
$$\mathbf{A} \in \mathbb{C}^{m\times n}_{\rho},
\qquad b \in \mathbb{C}^{m},
\qquad x \in \mathbb{C}^{n}$$
In short, when the data vector $b$ is not in the null space:
$$
b\notin\color{red}{\mathcal{N}\left( \mathbf{A}^{*} \right)}
$$
The least squares solution is the projection of the data vector $b$ onto the range space $\color{blue}{\mathcal{R}\left( \mathbf{A}\right)}$. A projection will exist provided that some component of the data vector is in the range space.
This post works through a derivation of the least squares solution based upon the singular value decomposition of a nonzero matrix.
How does the SVD solve the least squares problem?
Another approach is to start with the guaranteed existence of the singular value decomposition
$$
\mathbf{A} = \mathbf{U} \, \Sigma \, \mathbf{V}^{*},
$$
and build the Moore-Penrose pseudoinverse
$$\mathbf{A}^{+} = \mathbf{V} \Sigma^{+} \mathbf{U}^{*}.$$ The pseudoinverse is a matrix projection operator onto the range space $\color{blue}{\mathcal{R}\left( \mathbf{A}\right)}$.
For the full rank least squares problem, where $A \in \mathbb{K}^{m \times n},m>n=\mathrm{rank}(A)$ ($\mathbb{K}$ is the base field), the solution is $(A^T A)^{-1} A^T b$. This is a very bad way to approach the problem numerically for condition number reasons: you roughly square the condition number, so a relatively tractable problem with $\kappa=10^8$ becomes a hopelessly intractable problem with $\kappa=10^{16}$ (where we think about tractability in double precision floating point). The condition number also enters into convergence rates for certain iterative methods, so such methods often perform poorly for the normal equations.
The SVD pseudoinverse is exactly the same as the normal equations pseudoinverse i.e. $(A^T A)^{-1} A^T$. You simply compute it using the SVD and simplify. There is indeed a simplification; the end result is
$$(A^T A)^{-1} A^T = V (\Sigma^T \Sigma)^{-1} \Sigma^T V^T.$$
This means that if I know the matrix of right singular vectors $V$, then I can transform the problem of finding the pseudoinverse of $A$ to the (trivial) problem of finding the pseudoinverse of $\Sigma$.
The above is for the full rank problem. For the rank deficient problem with $m>n>\mathrm{rank}(A)$, the LS solution is not unique; in particular, $A^T A$ is not invertible. The usual choice is to choose the solution of minimal Euclidean norm (I don't really know exactly why people do this, but you do need some criterion). It turns out that the SVD pseudoinverse gives you this minimal norm solution. Note that the SVD pseudoinverse still makes sense here, although it does not take the form I wrote above since $\Sigma^T \Sigma$ is no longer invertible either. But you still obtain it in basically the same way (invert the nonzero singular values, leave the zeros alone).
One nice thing about considering the rank-deficient problem is that even in the full rank case, if $A$ has some singular value "gap", one can forget about the singular values below this gap and obtain a good approximate solution to the full rank least squares problem. The SVD is the ideal method for elucidating this.
The homogeneous problem is sort of unrelated to least squares, it is really an eigenvector problem which should be understood using different methods entirely.
Finally a fourth comment, not directly related to your three questions: in reasonably small problems, there isn't much reason to do the SVD. You still should not use the normal equations, but the QR decomposition will do the job just as well and it will terminate in an amount of time that you can know in advance.
Best Answer
The statement needs a crisp definition. Consider the matrix $\mathbf{A}\in\mathbb{C}^{m\times n}_{\rho}$ ($m$ rows, $n$ columns, rank $\rho$).
Uniqueness
If the data vector $b$ is not in the null space $\mathcal{N}(\mathbf{A}^{*})$ the least squares solution exists. If the number of columns is greater or equal to the rank of $\mathbf{A}$, $n\ge \rho$, the solution is unique.
Demonstration
The linear system $\mathbf{A}x=b$, $$ \begin{bmatrix}1 & 0\end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}b\end{bmatrix}, $$ offers the least squares solution $$ x_{LS} = \color{blue}{x_{particular}} + \color{red}{x_{homogeneous}} = \color{blue}{\begin{bmatrix}b \\ 0\end{bmatrix}} + \alpha \color{red}{\begin{bmatrix}0 \\ 1\end{bmatrix}} $$ where the arbitrary constant $\alpha \in \mathbb{C}$
The particular solution lives in the range space $\color{blue}{\mathcal{R}(\mathbf{A})}$. The homogeneous solution inhabits the null space $\color{red}{\mathcal{N}(\mathbf{A}^{*})}$. When this null space is trivial (when $n\ge \rho$), the is no homogeneous contribution and the least squares solution is unique.
Unique least square solutions