Categoricity is absolute.
By the Ryll-Nardzewski theorem, for a countable language, $\aleph_0$-categoricity of a complete theory $T$ is equivalent to $T$ proving for each natural number $n$ that there are only finitely many inequivalent formulas in $n$ variables. This property is evidently arithmetic and, thus, absolute.
Likewise, again in a countable language, it follows from the Baldwin-Lachlan theorem that a theory is categorical in some (hence, by Morley's theorem, all) uncountable cardinality just in case every model is prime and minimal over a strongly minimal set. Moreover, the strongly minimal formula may be taken to be defined over the prime model and the primality and minimality of every model over this strongly minimal formula is something which will be witnessed by an explicit analysis, hence, something arithmetic and absolute.
For uncountable languages, the situation is a little more complicated, but again categoricity is equivalent to an absolute property. Shelah shows that either the theory is totally transcendental and Morley's analysis in the case of countable languages applies, or the theory is strictly superstable though unidimensional.
Yes, all those assertions are expressible in the language of real closed fields, and all such assertions are decidable by Tarski's theorem. You can refer to addition, subtraction, multiplication, division, etc., plus the order; you can refer to any specific rational number (such as .223 etc.) because you have $1$ in the language and can therefore form any rational number; you can form any polynomial and indeed any rational function, with rational coeffficients, in any specific finite number of variables; and you can quantify over the real numbers and apply any kind of logical connective.
What Tarski proved is that in fact every such assertion is equivalent to a (much longer) quantifier-free assertion, and this is ultimately why they are decidable.
Theorem. (Tarski) The first-order theory of real-closed fields, which is the same as the theory of the structure $\langle\mathbb{R},+,\cdot,0,1,\lt\rangle$ is complete and decidable, and admits quantifier-elimination.
What you cannot do while remaining under the decidability result is quantify over the integers or the rational numbers. So you cannot necessarily decide questions like, "does this specific polynomial $p(\vec x)$ have an integer solution?". In particular, if you add a predicate for the integers to the structure, forming $\langle\mathbb{R},+,\cdot,\mathbb{Z},0,1,\lt\rangle$, where $\mathbb{Z}$ is a predicate that is true of exactly the real numbers that are integers, then the theory is no longer decidable. For this reason, the theory of the structure $\langle\mathbb{R},+,\cdot,\sin(x),0,1,\lt\rangle$ is not decidable, because in it we can define the integers. So while you can refer to specific polynomial or rational functions over the integers, you may not quantify over the class of such polynomials.
Similarly, you also cannot quantify over the dimension (which amounts to an integer quantifier), and ask something like: does every upper triangular $n\times n$ matrix (for every $n\geq 1$) have a certain property? That is, in order to stay within Tarski's language, you can in effect quantify over $\mathbb{R}^3$ or $\mathbb{R}^{86}$ or $\mathbb{R}^n$ for a specific $n$ by using $n$ individual real quantifiers, but you cannot treat $n$ itself as a variable here.
My opinion---probably overblown---is that Tarski's theorem is one of the great pinnacles of mathematical achievement. One of the consequences of his theorem, for example, is that Cartesian geometry is decidable. Tarski's theorem provides an algorithm that will decide by rote any question you can put to it about lines, circles, ellipses, cones, spheres, ellisoids, planes, cubes and so on, in any specific dimension. All such questions are expressible in the language of real-closed fields. (OK, I admit that the algorithm takes double exponential time, and is not feasible. But still, I find the existence of such an algorithm for a topic that has been studied by mathematicians for thousands of years to be amazing.)
Best Answer
Here is the outline of a proof that $Eq(\mathbb{R},+)$ proves all the equalities in both $Eq(\mathbb{R},+,\sin)$ and $Eq(\mathbb{R},+,\exp)$.
Step 1, defining $<$:
For any term $t$ in $L(+,\sin)$, define a real function $t_R$ by $$t_R(r_1,\ldots r_n)=\sup(\{|t(z_1,\ldots,z_n)|: z_1,\ldots,z_n \in \mathbb{C}, |z_1|=r_1, \cdots, |z_n|=r_n\})$$
Then for terms $t$ and $u$ in $L(+,\sin)$, define $t<u$ iff $$\mathbb{R}\models Qr_1 \ldots Qr_n\, t_R<u_R$$ where $Q$ is the quantifier "for all sufficiently large" and $n$ is the largest index of a primitive variable which appears in $t+u$.
Similary, for terms $t$ and $u$ in $L(+,\exp)$, define $t<u$ iff $$\mathbb{R}\models Qx_1 \ldots Qx_n\, t<u$$
Step 2, determining $<$:
I claim that for the definitions above:
Furthermore, this provides a recursive algorithm for determining whether $t<u$ for any terms $t$ and $u$ in the same one of the languages above.
Step 3, normalizing $<$:
For any term $t$ in the language, we get a normal form $N(t)$ by repeatedly applying the following rules:
I claim that $t<u$ iff $N(t)$ is smaller than $N(u)$ in the first part that differs.
Now, lexicographically, for any terms $t$ and $u$, either $N(t)<N(u)$ or $N(t)=N(u)$ or $N(t)>N(u)$. In the first and last cases, $t$ and $u$ are not equal over $\mathbb{R}$; for $\sin$ the definition of $<$ tells us that they are different over $\mathbb{C}$, and then the identity theorem tells us that they are also different over $\mathbb{R}$. In the middle case, $Eq(\mathbb{R},+)$ is enough to establish $t=N(t)=N(u)=u$.