Some hints on how I would proceed to do these:
For the first one, notice that in the theory of $({\bf Q},0,1,+)$ defines each rational number. If you add $<$ to this, you can find a distinct type for each real number (generally, if you have a linear order $(L,<)$, then with a parameters from a subset $A\subseteq L$ you can find distinct type for each element of the Dedekind completion of $A$).
In the second case, you can notice that the theory can be seen as the theory of vector fields over $\bf Q$, with added a symbol for one nonzero vector (because multiplication by rational numbers is definable in the theory). Looking at it this way, it shouldn't be hard to see that there are types for: 0, each element in the linear span of $1$ (which is countable), and for an element linearly independent with $1$ (as would be the case with every vector space over an infinite field with a symbol for one nonzero element). Alternatively, you can notice that there's a saturated countable model (${\bf Q}^\omega$).
In the third case, notice that for countable dense linear order, there's an automorphism of it which moves any element onto any other (which is easy to see). With little effort, using that fact you should be able to show that you can move any element in $({\bf Q},0,1,<)$ onto any other as long as you don't change its relationship with $0$ and $1$, so $({\bf Q},0,1,<)$ realizes exactly $5$ one-types; and if a model realizes only finitely many types, there can't be any more.
In the fourth case for $1$-types we can do something similar: show that for any two rationals $q_1,q_2\in \bf Q$ there's an automorphism that moves $q_1$ onto $q_2$ if and only if they are both positive, both negative, or both zero. For $2$-types you can do the same thing as in the first example (you can define the set of pairs $(q_1,q_2)$ such that $q_2$ is a given rational multiple of $q_1$, and using that, for each real number $r$, you can define the type which represents the pairs $(x,y)$ where $y=r\cdot x$).