$\mathbb{N}^*$ is countably infinite, so it's cardinality is equal to the cardinality of $\mathbb{N}$. The reason for this is that $\mathbb{N}^*$ is a countable union of countable sets. More generally, we may consider any countably or uncountably infinite set $X$. If $X^*$ consists of finite sequences of elements from $X$, then the cardinality of $X^*$ is equal to the cardinality of $X$. See: Countable_Union_of_Countable_Sets_is_Countable.
For $\mathbb{R}^\mathbb{N}$, its cardinality is equal to the reals, which says that there are as many infinite sequences of real numbers as there are real numbers. Details: cardinality-of-all-real-sequences.
First of all, let me clear a confusion of yours.
$\Bbb R$ has cardinality $2^{\aleph_0}$. This is not necessarily $\aleph_1$. It is consistent with the axioms of modern set theory that the answer is positive or negative. Similarly $2^{2^{\aleph_0}}$ need not be equal to $\aleph_2$, and it could be much larger.
Now that we cleared that issue. What is a number system? If you mean a field, simply something which has a structure of addition and multiplication like we know on $\Bbb Q$ or $\Bbb R$ or so on, then take any set $X$, and consider the ring of polynomials $\Bbb R[X]$ whose indeterminate variables come from $X$. Then $\Bbb R[X]$ is an integral domain, and therefore it has a fraction field, which is both a field and has the cardinality of $\max\{|\Bbb R|,|X|\}$. Picking a suitable $X$ would do.
If by a number system you mean something like ordinals, which are numbers (at least by virtue of being called "ordinal numbers") then ordinals have four natural operations, each defined from the previous one by transfinite induction: successorship, addition, multiplication and exponentiation.
Moreover we can show that if $\alpha$ and $\beta$ are infinite ordinals, then regardless to whatever operation you use on $\alpha$ and $\beta$, the cardinality of the result is the maximal one between $\alpha$ and $\beta$. This is wonderful, since it means that we don't have to look very far to find ordinals $\gamma$ with the property that $\{\alpha\mid\alpha<\gamma\}$ is closed under all operations.
For example, every infinite ordinal which is an initial ordinal (the least one of a given cardinality) is closed under all these operations. So $\omega_1$ and $\omega_2$ and $\omega_{\omega_1^\omega\cdot 5+\omega^{42}+1}$ are all such ordinals (recall that $\omega_\alpha$ denotes the $\alpha$-th initial ordinal). Pick any such ordinal whose cardinality is larger than the cardinality of the continuum, and you have an example.
It might be worth noting that the proper class of all ordinals is another such example, and it's not even a set!
On the other side of set theoretic number systems we have the cardinals, which again are called cardinal numbers, and these represent the size of a set. They also have the same four operations as ordinals, but now we can't define them from one another. We say that a cardinal $\kappa$ is a strong limit cardinal if whenever $\lambda<\kappa$ we have that $2^\lambda<\kappa$. This implies, amongst other things, that if $\kappa$ is a strong limit cardinal then we can show that the set of all cardinals below $\kappa$ (including the finite ones, of course) is closed under all operations on the cardinals, which will also make a number system.
And again, it might be worth noting that the class of all cardinals (what Cantor called Tav, the last letter of the Hebrew alphabet) is an example similar to the class of ordinals. It's a number system which is too large to even be a set.
Best Answer
You use the phrase "cardinalities greater than $C$," so I assume you know that Cantor's diagonal argument shows that for any set $X$, $\mathcal{P}(X)$ is strictly larger than $X$ (in that $X$ injects into $\mathcal{P}(X)$ but does not surject onto $\mathcal{P}(X)$).
Based on this, I think the real question is: what do you mean by "more uncountable"?
One possible answer is the following: there may be sets with combinatorial properties which are characteristic of extremely large objects, which "reasonable" infinite sets like the naturals and the reals cannot have. For example, measurability: a cardinal $\kappa$ is measurable iff there is a countably complete ultrafilter on $\kappa$ which is not principal. By combining "countably complete" with "nonprincipal," this is clearly an "uncountability property" if anything is!
I would guess that you would find large cardinals very interesting; and, I suspect that in general large cardinal properties provide positive answers to your question.
For a related question - given a set $X$ with cardinality in between $\omega$ and $C$, is $X$ "closer" to $\omega$ or $C$? - you should check out cardinal characteristics of the continuum.