In my elementary linear algebra class we did many problems like this without the Jordan Normal Form, etc, which I picked up later. So I will try to give a simpler explanation.
a) Recall that a matrix is non-invertible if it has nonzero vectors in its null space. This also means that those vectors are eigenvectors corresponding to the eigenvalue $0$. So what if $\text{null } A =\text{span }{(1,0,0)}?$ This is the case where an eigenvalue of $A$ is 0. However, you CANNOT conclude this from your givens, because you have no hypothesis on what the eigenvalue of $\text{span }{(1,0,0)}$ is.
b) Now, $A$ is a linear operator on $\mathbb R^3.$ This means that $A$ has a real eigenvalue, because the characteristic equation of $A$ is a cubic, so it has a real root. What if two of those roots were imaginary? Then $A$ would have its one-dimensional eigenspace, as given, but would not have a repeated eigenvalue.
c) This one is true. A matrix is diagonalizable if and only if has $n$ linearly independent eigenvectors, where $n$ is the dimension of the vector space it operates on. This means you need $3$ linearly independent eigenvectors where you only have one! An easy to remember this is that, to diagonalize a matrix $A$, you need to be able to write it as $A = PDP^{-1},$ where $P$ is an $n$ x $n$ matrix whose columns consist of eigenvectors of $A.$ Remember that $P$ is invertible if and only if it has linearly independent columns, and you are done.