= Solution
By <orthogonal diagonalization of a real symmetric matrix>, write $A=Q\operatorname{diag}(\lambda_1,\ldots,\lambda_d)Q^T$, with orthogonal $Q$. Its <matrix exponential> has the same eigenvectors and positive <eigenvalues> $e^{t\lambda_j}$. Orthogonal invariance of the induced <Euclidean norm> gives the exact identity
$$
\boxed{\|e^{tA}\|_2=\max_j e^{t\lambda_j}
=e^{t\lambda_{\max}(A)},\qquad t\geq0.}
$$
This proves the requested inequality with equality. If a real number $\beta$ gave the bound for every $t\geq0$, evaluating on a unit eigenvector for $\lambda_{\max}(A)$ at any $t>0$ would give $e^{t\lambda_{\max}}\leq e^{t\beta}$, so $\beta\geq\lambda_{\max}$. Thus \b[the stated exponent is the smallest possible]. For a <symmetric matrix> the <spectral abscissa> and <Euclidean logarithmic norm> coincide, unlike the general nonsymmetric case in Question 1.
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