What you say is correct and is true in more generality. You should be interested by the following result:
Let $K$ be a topological compact space. Then, the maximal ideals of $A
:= \mathcal{C}(K, \mathbb{R})$ are exactly the ideals of functions
vanishing at a fixed point of $K$.
Here are some hints to prove this theorem:
1) Determine the invertibles of $A$ (easy).
2) For any $x$ in $K$, show that the set of functions of $A$ vanishing in $x$ is a maximal ideal of $A$ (easy).
3) Conversely, let $I$ be a maximal ideal of $A$. Suppose by contradiction that for any $x$ in $K$, you can find a function $f_x$ in $A$ which does not vanish at $x$. By continuity, there is a neighborhood $U_x$ of $x$ such that $f_x$ does not vanish on $U_x$.
4) Using compactness, construct a function of $A$ that vanishes nowhere and conclude (clever). Hint: in the real numbers, a sum $\sum \lambda_i^2$ vanishes if, and only if, all $\lambda_i$ vanish.
Remark: adapt the proof, for functions with values in $\mathbb{C}$.
Edit: concerning the question of primality of ideals of functions vanishing at more than one point, you can prove this as follows:
Let $I_C$ be the set of functions of $A$ vanishing one some closed set $C$ of $[0,1]$. Write $C = X \cup Y$ where $X$ and $Y$ are closed proper subsets of $[0,1]$ in $C$ (this is always possible if $C$ has at least two points x < y: take for $X$ the set of points $t \in A$ with $t \leq \frac{x+y}{2}$ and for $Y$ the set of points $t \in A$ with $t \geq \frac{x+y}{2}$).
Let $f$ and $g$ be functions in $A$ vanishing exactly on $X$ and $Y$. For instance, you can take the distance functions to $X$ and $Y$. Then $f.g$ is in $I_C$ but neither $f$ nor $g$ are in $I_C$.
This can certainly be generalized for a more general topological space, but maybe the compactness condition is not sufficient.
You shouldn't have any trouble showing that your ring is isomorphic to $\mathbb Z_2^{10}$ (the product ring of $10$ copies of $\mathbb Z_2$.) The idea is that you send $\phi :\{1,2,\ldots, 10\}\to\mathbb Z_2$ to $(\phi(1),\phi(2),\ldots,\phi(10))\in \mathbb Z_2^{10}$.
Of course $\mathbb Z_2^{10}$ is a boolean ring since it is a product of boolean rings. This takes care of $(4)$.
It is well-known and easy to prove that the prime ideals are maximal in any boolean ring. Of course, it's also well-known and easy to prove that the prime ideals are maximal in finite rings. This takes care of $(2)$.
The ideals of a finite product of rings are easy to describe: they are all possible products of the ideals in the component rings. Since you have $10$ component rings, each of which have $2$ ideals, you have $2^{10}$ possible ideals. The whole ring is the one non-proper ideal you'll have to discard. This takes care of $(3)$.
The maximal ideals of a finite product of rings are also easy to describe: they are all of the form of the ideals described above, of course. It is easy to show that they are exactly those where you have a maximal ideal in one position, and the full ring in all other positions. So that gives you exactly $10$ maximal ideals, one for each position. This takes care of $(1)$.
Best Answer
a),b) and d) have been treated in the comments. For c) you should notice that the constant functions give you a natural inclusion $\mathbb R^* \subset C[0,1]^*$ and $\mathbb R^* = \mathbb R \setminus \{0\}$ is not even countable, in particular not cyclic.