For any square linear system $\,A\vec x=\vec b\,$ over some field, there exists a unique solution iff $\,\det A\neq 0\,$ , as then we can use the inverse matrix:
$$A\vec x=\vec b\Longleftrightarrow A^{-1}A\vec x=A^{-1}\vec b\Longleftrightarrow A^{-1}\vec b=\vec x $$
As for (a) and your "main question": if $\,\det A=0\,$ one still may have to check whether there are no solutions or infinite solutions (assuming we're working on an infinite field). For example, if the system is homogeneous (over an infinite field) it must have infinite solutions, whereas if the system is non-homogeneous it may have no solutions or several:
$$\begin{cases}x+y=1\\x+y=1\end{cases} \Longleftrightarrow \begin{pmatrix}1&1\\1&1\end{pmatrix}\binom{x}{y}=\binom{1}{1}\longrightarrow\,\,\text{infinite solutions}$$
$$\begin{cases}x+y=1\\x+y=0\end{cases} \Longleftrightarrow \begin{pmatrix}1&1\\1&1\end{pmatrix}\binom{x}{y}=\binom{1}{0}\longrightarrow\,\,\text{no solutions at all}$$
and, of course, in both cases above we have $\,\det A=0\,$
1st question: This means that the system $Bx=0$ has non trivial solutions (Why is that so? An explanation would be very much appreciated!).
If one of the rows of the matrix $B$ consists of all zeros then in fact you will have infinitely many solutions to the system $Bx=0$. As a simple case consider the matrix $M=\left(\begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right)$. Then the system $Mx=0$ has infinitely many solutions, namely all points on the line $x+y=0$.
2nd question: This is also true for the equivalent system $Ax=0$ and this means that A is non invertible (An explanation how they make this conclusion would also be much appreciated).
Since the system $Ax=0$ is equivalent to the system $Bx=0$ which has non-trivial solutions, $A$ cannot be invertible. If it were then we could solve for $x$ by multiplying both sides of $Ax=0$ by $A^{-1}$ to get $x=0$, contradicting the fact that the system has non-trivial solutions.
Best Answer
The system $Ax=0$ always has the trivial solution, and $Ax=b$ when $b≠0$ does not. Having an infinite number of solutions does not necessarily mean that $0$ is one of them; consider the system:
$A=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, $b=[1,0]$
Every $x=[y,1]$ (for every $y$) solves $Ax=b$, thus you have infinite solutions. However $x=[0,0]$ is not a solution.