Remember the definition of bounded: A function $h\colon\mathbb{R}\to\mathbb{R}$ is bounded if and only if there exist number $M,N\gt 0$ such that for all $r\in\mathbb{R}$, $-N \leq f(r) \leq M$.
Intuitively, the graph of $f$ goes neither "too high" nor "too low".
So, suppose that $f$ and $g$ are both bounded. That means that there the values of $f$ don't get bigger than some $M_1\gt 0$, and the values of $g$ don't get bigger than some $M_2\gt 0$. When you take $(f+g)(r) = f(r)+g(r)$, how big can the number $f(r)+g(r)$ be, given that $f(r)\leq M_1$ and $g(r)\leq M_2$?
How about discussing how small they can be, given that you know there are numbers $N_1$ and $N_2$ such that $f(r)\geq N_1$ and $g(r)\geq N_2$?
How about the product? That's a bit harder because of the signs, so try working with $|fg|$, $|f|$, and $|g|$. Notice that if $-N\leq f(r)\leq M$, then $0\leq f(r) \leq \max\{M,N\}$ (verify this!).
Number $3$ is a bit trickier... first question: can the sum of two functions, each of which is not bounded, be bounded? Yes: consider for instance $f(x) = x$, which is not bounded, and $g(x) = 1-x$, which is also not bounded. What is $f+g$? Of course, this example does not satisfy that $fg$ is bounded (here, $fg(x) = x-x^2$, which is unbounded). And it's easy to come up with unbounded examples with their product bounded (just make sure that whenever $f$ gets out of control, $g$ "cancels it out").
Perhaps you may want to check that if $ff$ (the product of $f$ with itself) is bounded, then so is $f$, and then consider Theo Buehler's answer.