= Solution
\b[(1) True.] Over $\mathbb C$, $A$ has an <eigenvalue> $\lambda$ with nonzero <eigenspace> $E$. Commutativity makes $E$ invariant under $B$: $A(Bv)=B(Av)=\lambda Bv$. The <endomorphism> $B|_E$ has an <eigenvector> $v\ne0$ because its <characteristic polynomial> has a complex root. This $v$ is an <eigenvector> of both matrices, though their <eigenvalues> on it may differ.
\b[(2) False.] Take $A=I$ and $B=2I$. They commute, but their only <eigenvalues> are respectively $1$ and $2$.
\b[(3) True.] Associativity gives $(BA)^{n+1}=B(AB)^nA=0$, so $T=BA$ is nilpotent. A nilpotent <endomorphism> of an $n$-dimensional space has index at most $n$: in $V\supseteq TV\supseteq T^2V\supseteq\cdots$, equality at a nonzero image would make $T$ surjective on that finite-dimensional image, hence invertible there, contradicting nilpotence. Thus dimensions strictly decrease until zero, in at most $n$ steps. Therefore $(BA)^n=0$.
\b[(4) False.] Take $T=0$ on a two-dimensional <vector space> and $\lambda=0$. The <eigenspace> has dimension two, but the <minimal polynomial> is $t$, in which the root zero has multiplicity one. <Eigenspace> dimension counts independent <eigenvectors>; the exponent in the <minimal polynomial> measures the longest corresponding <Jordan chain>, not that dimension.
\b[(5) True.] Factor the <minimal polynomial> as $m_T(t)=(t-\lambda)^c q(t)$, with $q(\lambda)\ne0$, and set $S=T-\lambda I$. If $v\in W_{c+1}$, then $w=S^cv$ satisfies $Sw=0$, so $q(T)w=q(\lambda)w$. But $m_T(T)v=0$ also gives $q(T)w=0$. Thus $w=0$, so $v\in W_c$. The reverse inclusion is immediate from $S^cv=0\Rightarrow S^{c+1}v=0$, proving
$$
\boxed{W_c=W_{c+1}.}
$$
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