A function $f:\mathbb R\to\mathbb R$ is Borel if and only if for all $a\in\mathbb R$, the set $\{x\in\mathbb R:f(x)>a\}$ is a Borel subset of $\mathbb R$.
Suggestion: Think about what possible types of sets you can get for $\{x\in\mathbb R:f(x)>a\}$ when $f$ is a monotone function. You may want to conjecture with the aid of examples before trying to prove your conjecture. The sets you get should be easily confirmed to be Borel.
The answer to #1 is yes.
First note that if $f$ is monotone, it is Borel. (The sets $(a, \infty)$ generate the Borel $\sigma$-algebra, and $f^{-1}((a, \infty))$ is Borel for each $a$ because it is of the form $(b, \infty)$ or $[b, \infty)$ (for increasing functions) or $(-\infty, b)$ or $(-\infty, b]$ (for decreasing functions).)
Now for each $y$, $f^{-1}(\{y\})$ is either empty, a point, or a nontrivial interval. Let $C$ be the set of all $y$ such that $f^{-1}(y)$ is a nontrivial interval. Since each interval contains a rational, $C$ is countable. Let $D = f^{-1}(C)$; note that $D$ is Borel.
If $B$ is an arbitrary Borel set, we have $f(B) = f(B \cap D) \cup f(B \setminus D)$. Now $f(B \cap D) \subset C$, hence it is countable and hence Borel. So it suffices to show that $f(B \setminus D)$ is Borel.
On $D^c$, and hence on $B \setminus D$, $f$ is injective. Now it is a theorem of descriptive set theory that an injective Borel function on a Borel subset of a Polish space has a Borel image. (See for instance Theorem 4.5.4 of Srivastava, A Course on Borel Sets.) But $B \setminus D$ is Borel, so $f(B \setminus D)$ is also Borel and we are done.
Best Answer
Hint: If $f$ is monotone, then, for every real number $x$, the set $$f^{-1}((-\infty,x])=\{t\mid f(t)\leqslant x\}$$ is either $\varnothing$ or $(-\infty,+\infty)$ or $(-\infty,z)$ or $(-\infty,z]$ or $(z,+\infty)$ or $[z,+\infty)$ for some real number $z$.
To show this, assume for example that $f$ is nondecreasing and that $u$ is in $f^{-1}((-\infty,x])$, then show that, for every $v\leqslant u$, $v$ is also in $f^{-1}((-\infty,x])$.