I will go with the following solution:
We can write
\begin{align*}
\Gamma_A &= \bigcap_{x,y\in A} \{ \varphi \in A' : \varphi(xy) = \varphi(x) \varphi(y) \} \cap \{\varphi: \varphi(1_A) =1 \} \\
&= \bigcap_{x,y \in A} \{ \varphi \in A': (xy)'(\varphi) - x'(\varphi) y'(\varphi) = 0 \} \cap \{\varphi: 1_A'(\varphi)=1 \}.
\end{align*}
Now, notice that $x'$, $y'$, $(xy)'$ and $1_A'$ are (by definition) continuous functions from $A'$ to $\mathbb{K}$. Since
$$
+ : \mathbb{K} \times \mathbb{K} \to \mathbb{K},\ (x,y)\mapsto x+y \quad \text{and} \quad \cdot: \mathbb{K} \times \mathbb{K} \to \mathbb{K},\ (x,y)\mapsto x \cdot y
$$
are continuous functions, we know that $(xy)' - x' \cdot y'$ is also a continuous function.
Thus, $\Gamma_A$ is closed as the intersection of closed sets.
I think that your first question is very broad and can have many interesting answers depending on the context. The point is that, since you are considering $C(X,Y)$, the answer depends a lot on what is the topology of the other space $Y$.
If you allow the topology of $Y$ to vary too, you can essentially make the set $C(X,Y)$ to be either the set of constant functions or the set of all functions. But if the topology of $Y$ is fixed (and somewhat nontrivial) you could have interesting behaviours, see the answer by Jochen.
Concerning the second question, as (again) Jochen suggests in a few comments, you can construct many examples, e.g. by composing any linear functional $f\in X^*$ with any continuous function $\phi:\mathbb R\to\mathbb R$.
In the above example, though, you are basically reducing yourself to the finite-dimensional case, where weak and strong topologies coincide. In fact, you can view any non-trivial functional $\phi\in X^*$ as a linear projection onto a one-dimensional space. I think that constructing a "genuinely infinite-dimensional" example is a bit less trivial (if not impossible, I don't know).
Even though this is not completely related, I want to mention the case of convex functions: if you have a continuous convex functional $\phi:X\to\mathbb R$, this is authomatically lower-semicontinuous in the weak topology of $X$ (this is related to the fact that a strongly closed, convex subset of a Banach space $X$ is also weakly closed), but you cannot upgrade this statement to continuous functions. In fact (again, again), Jochen suggests the example of the norm $||\cdot\||$, which is convex and continuous, thus weakly lower-semicontinuous, but not weakly continuous (there exist sequences of unitary vectors that converge weakly to zero). If you want to look at some references, have a look at the book of H. Brezis "Functional Analysis, Sobolev Spaces and Partial Differential Equations".
Best Answer
These two things do not really mix much. The weak topology is defined on a topological vector space: you need a vector space to be able to define linear functionals. The weak$^*$ topology makes sense only on duals.
The Gelfand topology is used to give a topology to the set of characters of an abelian C$^*$-algebra $A$ (it can be done for a Banach algebra, too). What you do is you consider the characters as a subset $\Sigma$ (not a subspace!) of the dual $A^*$ of $A$, and you endow $\Sigma$ with the relative weak$^*$-topology (which is simply pointwise convergence).
The weak$^*$ topology is nice because it makes closed balls compact, which is often very useful. In particular it makes $\Sigma$ above compact when it's closed (which is precisely when $A$ is unital), and locally compact in general.