Suppose $H$ is a separable Hilbert space and $F:H \to \mathbb{R}$ is a map (continuous if necessary). Let $K= \{a_n\}\cup \{a\}$ where $a_n \to a$ in $H$.
Isn't it always the case that $$\sup_{h \in K} F(h) = \begin{cases} F(a_{k_*}) &: \text{for some $k_* \in \mathbb{N}$}\\ F(a) &: \text{o/w} \end{cases}$$ and hence, if the $F$ depends on a parameter $p$, $$\lim_{p \to p_0} \sup_{h \in K} F(h) = \begin{cases} \lim_{p \to p_0}F(a_{k_*}) &: \text{for some $k_* \in \mathbb{N}$}\\ \lim_{p \to p_0}F(a) &: \text{o/w} \end{cases}$$ Is it correct?
A convergent sequence together with its limit is always a compact space (at least in Hausdorff spaces). So the answer to your first question is "yes", because $F(K)$ is compact in $\mathbb{R}$. That's under assumption that $F$ is continous. Otherwise this is not true, $F(K)$ can be any countable subset of $\mathbb{R}$ (doesn't have to be closed, so $\sup$ doesn't have to belong to it).
The answer to the second question is "no", even when $F$ is continous and $H$ is well behaving space like $\mathbb{R}$. That's because the limit $\lim\sup F_p(K)$ doesn't even have to exist. For example take $a_n=\frac{1}{n}\in\mathbb{R}$. Then
$$K=\bigg\{0, 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots\bigg\}$$
Define
$$F_p:\mathbb{R}\to\mathbb{R}$$ $$F_p(x)=(-1)^p\cdot x$$
Then
$$\sup F_p(K)=\begin{cases} 0 &\mbox{ for odd } p \\ 1 &\mbox{ for even } p \end{cases}$$
So the limit doesn't exist.