Here is an exercise in model theory.
- The theory of $(\Bbb{R}, 0, +)$ has exactly two $1$-types but $\aleph_0$ many $2$-types.
- The theory of $(\Bbb{R}, 0, +, <)$ has exactly three $1$-types but $2^{\aleph_0}$ many $2$-types.
For 1, two $1$-types are $x+0=x$ and $x+(-x)=0$. For $2$-types, I think one is $x+y=y+x$, but I am not sure how to get $\aleph_0$ many $2$-types.
For 2, three $1$-types are $x+0=x$, $x+(-x)=0$ and $x>0\lor x=0\lor -x>0$. But I am not sure how to get $2^{\aleph_0}$ many $2$-types.
Note that a type in a theory $T$ is a maximal consistent set $Γ(x_1,⋯,x_{n})$ of formulas in the variables $x_1,⋯,x_{n}$.
Whenever i say type, I mean complete type without parameters (as this is how you are intending them).
Hints
Reminder I: if $\mathcal{L}$ is a first order signature, $M$ is a $\mathcal{L}$-structure and $\overline{m} \in M^n$ is a $n$-tuple in $M$, then the set of $\varphi(\overline{x})$ such that $M \models \varphi(\overline{m})$ is a type, it is referred to as the type of $\overline{m}$.
Reminder II: and $(r_0, \ldots, r_{n-1})$ and $(s_0, \ldots, s_{n-1})$ are two $n$-tuples in a $\mathcal{L}$-structure $M$ that are are conjugated by a $\mathcal{L}$-automorphism of $M$, then they have the same type.
Ingredient I: if $r \in \mathbb{R}\setminus \{0\}$, then $x\mapsto rx$ is an automorphism of $(\mathbb{R}, 0, +)$; if $f: \mathbb{R} \rightarrow \mathbb{R}$ is any $\mathbb{Q}$-linear bijection, then it is an automorphism of $(\mathbb{R}, 0, +)$.
Ingredient II: if $r \in (0, \infty)$ then $x \mapsto rx$ is an automorphism of $(\mathbb{R}, 0, +, <)$.
Solutions
$1$-types of $(\mathbb{R}, +, 0)$
The set of consequences of $x\neq 0$ is a $1$-type: it is maximal as it can alternatively be construed as the type of a fixed element $r \in \mathbb{R} \setminus \{0\}$. The other $1$-type is $x=0$ (clearly maximal). These are the only types as if you $\Phi(x)$ is a type, then either $(x=0) \in \Phi(x)$ or $(x\neq 0) \in \Phi(x)$.
$2$-types of $(\mathbb{R}, +, 0)$
Using Ingredient I you can see that all pairs $(r,s) \in \{0\} \times (\mathbb{R} \setminus 0)$ have the same type generated by $\varphi_{0,1}(x,y)=(x=0 \land y \neq 0)$. Simiarly you can deduce that $\varphi_{1,0}(x,y)=(x\neq 0 \land y=0)$ and $\varphi_{0,0}(x, y)= (x=0 \land y=0)$ also generate complete types.
Given $q =\frac{m}{n} = \mathbb{Q}\setminus 0$ all $(r,s) \in (\mathbb{R}\setminus \{0\})^2$ such that $r/s=q$ all have the same type generated by $\varphi_q(x,y)= (nx = my \land x \neq 0)$. Finally note that all the pairs $(r,s) \in (\mathbb{R}\setminus \{0\})^2$ not standing in a rational ratio all belong to the same type because of the last part of Ingredient I. Their type is not principal; it is generated by the set of formulae $\Phi(x,y)=\{ (mx \neq ny): m,n \in \mathbb{Z}\}$.
These are all the $2$-types of $(\mathbb{R}, +, 0)$. Indeed let $\Psi(x,y)$ be any comlete type: either you have that one among $\varphi_{0,1}, \varphi_{0,0}, \varphi_{1,0}, \varphi_q$ ($q$ varying in $\mathbb{Q}$) belongs to $\Psi(x,y)$ and in such a case $\Psi$ is the types spanned by that formula, or $\Psi$ contains $\lnot\varphi_{0,1}, \lnot\varphi_{0,0}, \lnot\varphi_{1,0}, \lnot\varphi_q$ for every $q \in \mathbb{Q}^*$ and in such a case $\Psi(x,y)=\Phi(x,y)$.
$1$-types of $(\mathbb{R}, 0, +, <)$
You are correct on pointing out the types are the three generated by the formulae $(x=0)$, $(x>0)$ and $(x<0)$ respectively. You can prove these generate complete types using Ingredient II. To prove these are the only ones proceed similarly to what you did above.
$2$-types of $(\mathbb{R}, 0, +, <)$
You don't need to list all of them: you know there are at most $2^{\aleph_0}$ types because you are considering a countable language. Just note that if you have $(r,s)\in (0, \infty )^2$ and $(r',s')\in (0, \infty )^2$ such that $r/s < r'/s'$ then they have different types: you can find $q=\frac{m}{n}\in \mathbb{Q}$ such that $r/s< q<r'/s'$, then $(r,s)$ satisfies $\varphi_{q}^{<}(x,y)= (nx<my)$ whereas $(r',s')$ does not.