If $A$ is real symmetric matrix then there exists an orthogonal matrix $Q$ such that
$$ A = QDQ^T$$
Then, if $f$ is a polynomial, one can write
\begin{equation} f(A)=Qf(D)Q^T \end{equation}
and also $$\|f(A)\| = max_{\lambda \in \sigma(A)}|f(\lambda)|$$
where the last equation follows from the fact that $A$ is a symmetric matrix and hence $\rho(A) = \|A\|$
$\textbf{First Question}:$ To find out the third equation from the fact that $f(A) = Qf(D)Q^T$ , so $\|f(A)\| = \|Qf(D)Q^T\|$ is enought o say that an orthogonal matrix doesn't affect the norm and therefore $$\|Qf(D)Q^T\| = \|f(D)\|?$$
$\textbf{Second Question}:$ What if A is not symmetric but still diagonalizable, so
$$A = PDP^{-1}$$
with $P$ a matrix of eigenvectors, then for any polynomial I can still write
$$f(A) = Pf(A)P^{-1}$$
,but what can I say about $\|f(A)\|$? Only that, in general
$$\|f(A)\| \ge \rho(A)?$$
First Question : The frobenius norm of $f(A)$ is given by \begin{align*} \|f(A)\|_F^2 &= \sup_{x} \frac{\|f(A) x\|^2_2}{\|x\|_x^2}\\ &= \sup_x \frac{x^T Q f(D)^T f(D) Q^T x}{x^T x}\\ &= \sup_y \frac{y^T f(D)^T f(D) y}{y^T Q^T Q y}\\ &= \sup_y \frac{\|f(D) y\|^2}{\|y\|^2}\\ &= \|f(D)\|_F^2 \end{align*}
Second Question : The above result is true for any square matrix $A$, you can prove it using the Jordan Normal form, in this case $D$ is a Jordan matrix and then $\|f(A)\|_F = \|f(D)\| \geq \max_{\lambda\in\sigma(A)} |f(\lambda)|$. In general it is not true that $\|f(A)\|_f\geq \rho(A)$, imagine if $f(B)=0$ for all $B$ and $A$ is positive definite.
Observe that your first result also holds for any $C^\infty$ function, not only polynomials, since a $C^\infty$ is an infinite polynomial (Taylor development).