Your answer to part 1 is fine.
The notation means "functions from the interval $[0, 1]$ to the reals"; the interval $[0, 1]$ is the set of all real numbers between $0$ and $1$, inclusive.
For part 2, the notation $[\cos x]$ means "the equivalence class of the function $x \mapsto \cos x$. What functions are equivalent (under this relation) to cosine? Ones that have the same value at $x = 0$ as cosine, i.e., all functions $f$ with $f(0) = 1$.
You've observed that for each number $i \in \mathbb R$, there's an equivalence class (consisting of all functions that take the value $i$ at $x = 0$). So the set of equivalence classes is in 1-to-1 correspondence with the real numbers: the functions whose value at $0$ is $i$ correspond to the real number $i$.
For part 4, you need to find one element of each class. So a typical class is the one for $i = 3.5$. Can you think of a function whose value at $0$ is $3.5$? Sure, you can think of millions of them. But a particularly easy one is the constant function $f(x) = 3.5$. And indeed, the set of all constant functions has the property that each function is in a different equivalence class, and each equivalence class contains one of these functions.
You were 90% of the way there with your reasoning. Keep up the good work.
Here are two examples :
$1 - $ Consider the relation $\equiv$ ( an equivalent relation), then
$$a \sim b \Leftrightarrow a\equiv b \mod 2 $$
That is, $a$ and $b$ will be in the same class $\overline{a}$ if their remainders of the division by $2$ are the same. For example $4$ and $6$ belong to the same class, which we are going to choose a representant $0$, because
$$6 = 3 \cdot 2 + \color{red}{0} \ \ \text{and} \ \ 4 = 2 \cdot 2 + \color{red}{0}$$
then we say $\overline{4} = \overline{6} = \overline{0}$. If we think, there are two distinct classes: $$\overline{0} = \{x \in \mathbb Z ; x \equiv 0 \mod 2, \text{$x$ is even}\}\ \ \text{and}\ \ \overline{1} = \{x \in \mathbb Z ; x \equiv 1 \mod 2, \text{$x$ is odd}\}$$
The set of all classes is
$$\mathbb Z_2 = \{\overline{0}, \overline{1}\}$$
$2-$ Consider the relation
$$(a,b) \sim (c,d) \Leftrightarrow ac = bd $$
This equivalent relation gives us the fractions, that is the filed of fractions of $\mathbb Z$. Similarly we choose a class representant for example,
$$\frac{1}{2} = \frac{2}{4} = \frac{3}{6 } = \cdots$$
we choose $\frac{1}{2}$ to be the class representant. Notice that $\mathbb Q = \{ \frac{a}{b} ; a,b \in \mathbb Z, \ \ \text{where}\ \ b \neq 0\}$ is the set of all classes.
Best Answer
First note that equivalence classes $R$ are subsets of $A$. Because the size of any subset of a countable set is countable, all equivalence classes are countable.
For part $3$, suppose all equivalence classes are countable. The order of a union countable sets is also countable, so the order of $A$ (which is the order of the union of all equivalence classes) is countable. This is a contradiction, so there is at least one non-countable equivalence class.