(a) If $C=AB-BA$ we see that $\operatorname{tr} C = 0$. If $(I-C)^n=0$, then $I-C$ is nilpotent and all eigenvalues are zero. Hence $\operatorname{tr} (I-C) = 0$, which implies that $\operatorname{tr}C = n$, a contradiction. Hence no such matrix can exist.
(b) The arithmetic and geometric means are related by
${x_1+...+x_n \over n} \ge \sqrt[n]{x_1\cdots x_n}$, which is equivalent to
$(x_1+...+x_n)^n \ge n^n x_1\cdots x_n$. Since $\operatorname{tr} A = \lambda_1+\cdots + \lambda_n$ and $\det A = \lambda_1 \cdots \lambda_n$, we obtain the desired result.
(c) All eigenvalues of a skew symmetric matrix are imaginary. Since it is real and has odd dimension, it must have one real eigenvalue. The only number that is both real and imaginary is zero. Hence $A$ is singular.
You can use physical ideas to help construct a simple counterexample to show that unless additional conditions are imposed, like positive definiteness etc, the symmetric part can represent quite different things.
Consider (for simplicity) a 2D body (a unit square with one vertex at the origin) in a fixed cartesian basis, and subjected to the following homogeneous deformation.
$$
x = X + 2Y \\
y = X - Y
$$
This is not change of basis, but an actual change in the position / shape of the body, with lowercase denoting the "deformed" configuration and uppercase the coordinates of a point in the body in its original configuration.
In matrix terms
$$
\begin{bmatrix} x \\ y \end{bmatrix}
=
\begin{bmatrix}
1 & 2 \\
1 & -1 \\
\end{bmatrix}
\begin{bmatrix} X \\ Y \end{bmatrix}
$$
We'll call this matrix $F$ and write it as a sum of symmetric and skew-symmetric components
$$
F = F_{s} + F_{skew} \\
F = \frac{1}{2}(F + F^T) + \frac{1}{2}(F - F^T) \\
=
\begin{bmatrix}
1 & 3/2 \\
3/2 & -1 \\
\end{bmatrix}
+
\begin{bmatrix}
0 & 1/2 \\
-1/2 & 0 \\
\end{bmatrix}
$$
Now consider what the first (symmetric) matrix $F_s$ does to our unit square with vertices $a=(0,0), b=(1,0), c=(1,1) \; and \; d=(0,1)$ in that order. (We might as well start with symmetric F, but anyway).
The new coordinates of the unit square are (let's call them $a', b', c', d')$ under this symmetric map are
\begin{bmatrix}
a' & b' & c' & d' \\
0 & 1 & 5/2 & 3/2 \\
0 & 3/2 & 1/2 & -1 \\
\end{bmatrix}
Looks ok. However, if you sketch this figure you'll see that the handedness of the figure has flipped. In particular, its Jacobian is negative, so the area changes sign. What this implies is that no matter how you parametrize or decompose (say using products) the map $F_s$, at some stage, material in the square must interpenetrate to reach the final configuration. So in this instance, the symmetric matrix with no other restriction represents an impossible deformation (if you look at it physically). Mathematically, it represents a change of handedness by interchanging sides PLUS a stretch in this case.
Even weirder things can happen when the jacobian is 0. The symmetric map
\begin{bmatrix}
1 & -1 \\
-1 & 1 \\
\end{bmatrix}
results in collapsing our unit square into a straight line passing through (-1,1) and (1,-1) with obviously no area! So you see that symmetry by itself doesn't tell you what's going on geometrically unless you look at additional characteristics of the map.
Best Answer
The case of $A$ being symmetric fails to imply that $B$ is anti-symmetric in a pretty trivial case (let's ignore the zero matrix since we often ignore $0$ in many arguments like this): $A = I_{n\times n}$ as this only implies $B$ is traceless, nothing more.
The case of $A$ being anti-symmetric fails to imply that $B$ is symmetric in general as well but for slightly less trivial reasons. Note: if $A$ and $B$ are $2\times 2$ matrices, the anti-symmetry of $A$ does imply the symmetry of $B$. (Check it yourself.) To see that it fails in general we then need to consider $3\times 3$ matrices or larger. Let $A$ and $B$ be given as below.
$$A = \left(\begin{array}{rrr} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{array}\right)$$
$$B = \left(\begin{array}{rrr} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 0 & 1 & 0 \end{array}\right)$$
Then $\text{tr}(AB) = 0$ and $A$ is anti-symmetric but $B$ is not symmetric.