An $n$-type for a language $\mathscr{L}$ is a maximal consistent set of formulas of $\mathscr{L}$ with $n$ free variables, so a $1$-type is a maximal consistent set of formulas of $\mathscr{L}$ with one free variable. If $\mathfrak A$ is a model for $\mathscr{L}$ with underlying set $A$, and $a\in A$, the set of all formulas $\varphi(x)$ such that $\mathfrak A\vDash\varphi(a)$ is a $1$-type, called the type of $a$ in $\mathfrak A$.
Intuitively, a $1$-type says as much as one can consistently say (in $\mathscr{L}$) about a single object, so there’s a $1$-type for each distinguishable kind of object. Look at $\operatorname{Th}(\langle\Bbb Q,<,0,1\rangle)$, for instance; what kinds of rationals are distinguishable using only the relation $<$ and the constants $0$ and $1$? Clearly the type of $0$ is unique: the formula $x=0$ by itself picks out that one rational number, and expanding it to a $1$-type isn’t going to change that. The same goes for $1$, so we’ve found two $1$-types already. There are supposed to be three more, and it’s pretty obvious what they must be: the two named rationals divide the rest of $\Bbb Q$ into three intervals, $\Bbb Q\cap(\leftarrow,0)$, $\Bbb Q\cap(0,1)$, and $\Bbb Q\cap(1,\to)$. You can say that a rational is in one of these intervals: the formulas $x<0$, $0, and $1 already characterize these three types. On the other hand, there is no collection of formulas in one free variable that is satisfied by some of the negative rationals but not by others: there simply isn’t anything that you can say in this language that distinguishes one negative rational from another.
Now take a look at $\operatorname{Th}(\langle\Bbb Q,<,+\rangle)$; there are supposed to be three $1$-types and uncountably many $2$-types. Start with the $1$-types; are there any specific rationals that we can pin down in this language? There’s definitely one: the formula $x+x=x$ is satisfied only by $0$, so one of the $1$-types defines $0$. Since we have $<$, we can also say that a rational is positive or that it’s negative: the formula $\exists y(y+y=y\land y is satisfied precisely by the positive rationals. But where are all those $2$-types going to come from?
Start ‘small’: try to find a countably infinite number of different $2$-types. If $m$ and $n$ are positive integers, let $\varphi_{m,n}(x,y)$ be the formula
$\underbrace{x+\ldots+x}_m=\underbrace{y+\ldots+y}_n\;;$
$\varphi_{m,n}(x,y)$ characterizes pairs $\langle p,q\rangle$ of rationals such that $mp=nq$, i.e., such that $\frac{p}q=\frac{n}m$. This is almost like being able to specify each rational, and it gets us $\omega$ $2$-types. There are uncountably many Dedekind cuts in $\Bbb Q$; perhaps we should try to modify the preceding idea so that instead of picking out particular rational ratios, it picks out arbitrary real ratios. Let $\psi_{m,n}(x,y)$ be the formula
$\underbrace{x+\ldots+x}_m<\underbrace{y+\ldots+y}_n\;;$
$\psi_{m,n}(p,q)$ says that $mp, or $\frac{p}q<\frac{n}m$, and $\psi_{n,m}(q,p)$ says that $nq, or $\frac{n}m<\frac{p}q$. Using these formulas, you can construct sets of formulas that are satisfied by a pair $\langle p,q\rangle$ of rationals if and only if the ration $\frac{p}q$ satisfies an infinite family of inequalities of the forms $\frac{p}q<\frac{n}m$ and $\frac{n}m<\frac{p}q$; do you see how you can combine that with the idea of Dedekind cuts to generate an uncountable family of $2$-types?
I’ll say nothing about the other three problems; I think that I’ve given you enough here to give you a good start on tackling them.