= Solution
The <matrix exponential> acts entrywise on a diagonal matrix, so
$$
\boxed{e^B=\operatorname{diag}(e^3,e^4,1).}
$$
For the two-dimensional matrix $C$, direct multiplication gives $C^2=-C$, hence $C^n=(-1)^{n-1}C$ for $n\ge1$. Summing its defining <power series> gives
$$
\boxed{e^C=I+(1-e^{-1})C
=\begin{pmatrix}1&0\\1-e^{-1}&e^{-1}\end{pmatrix}.}
$$
Finally, $e^A=V\operatorname{diag}(e,e^2,e^5)V^{\mathsf T}$, so each diagonal entry is a weighted sum of these exponential eigenvalues, with weights equal to squared eigenvector components:
$$
\boxed{(e^A)_{11}=\frac{e^2}{3}+\frac{2e^5}{3},\qquad
(e^A)_{22}=(e^A)_{33}=\frac e2+\frac{e^2}{3}+\frac{e^5}{6}.}
$$
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