a) is correct
b) No, it's not true. If $A=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$, then the sum of the singular values is $2$, while the absolute value of the trace is zero.
To discuss whether your matrix is ill-conditioned, you need to say which norm you are talking about. Assuming we are talking about the operator norm (=largest singular value), if the determinant is small it means that some singular values are small; then the inverse will have big singular values and the condition number will be large.
c) You have $\sigma_1\sigma_2\cdots\sigma_n<10^{-k}$; if all $\sigma_j\geq 10^{-k/n}$, then $$|\det A|=\sigma_1\cdots\sigma_n\geq(10^{-k/n})^n=10^{-k};$$
so at least one singular value is less than $10^{-k/n}$.
2) This is not well phrased, because they can be equal. The maximum singular value is $\|A^TA\|^{1/2}$. Now let $\lambda$ be
an eigenvalue of $A$ with unit eigenvector $v$. Then
$$
|\lambda|=\|\lambda v\|=\|Av\|=(v^TA^TAv)^{1/2}\leq\|A^TA\|^{1/2}(v^Tv)^{1/2}=\|A^TA\|^{1/2}.
$$
So every eigenvalue is smaller in absolute value than the biggest singular value.
First, the difference in the eigenvectors. Let $(\lambda,v)$ be an eigenpair of $A$, i.e., $A v = \lambda v$ and let $\alpha \in \mathbb{C} \setminus \{0\}$. Then
$$A (\alpha v) = \alpha A v = \alpha \lambda v = \lambda (\alpha v).$$
So, $v$ is an eigenvector of $A$ if and only if $\alpha v$ is an eigenvector of $A$. Both are equally "good", unless you desire some additional properties. Note that this works for any $A$, not just $A = C$.
Second, the significance of the left singular vectors is in computing the eigenvalue decomposition in $XX^T$ (in your notation: $X^T = X'$).
Third, a real diagonal matrix is orthogonal if and only if each of its diagonal elements is either $1$ or $-1$. Let us prove this.
Let $D = \mathop{\rm diag}(d_1,\dots,d_n)$. Obviously, $D = D^T$, so
$$D^TD = \mathop{\rm diag}(d_1^2,\dots,d_n^2).$$
So, $D^TD = {\rm I}$ if and only if $d_k^2 = 1$ for all $k$.
For complex matrices (and using complex adjoint $Q^*$ instead of transpose $Q^T$), we get that $|d_k| = 1$ for all $k$.
Best Answer
Here's one way to interpret it "geometrically":
Consider the set $S = \{x \in \Bbb R^n: \|x\|_2 = 1\}$, which is the "$n$-dimensional hypersphere" of radius $1$. The map $A$ take the sphere $S$ to some "hyper-ellipsoid" $A(S)$.
The condition number is a measurement of how "skinny" the resulting ellipsoid is (specifically, it is the ratio of the lengths of the "major axis" and "minor axis"). The worse-conditioned the matrix, the "skinnier" the ellipsoid.
If $A$ is not invertible (i.e. if its condition number is $\infty$), then the resulting ellipsoid is "flat".