Are these implications about topology true or false?
All are false. For (1) let $f(x)=x^2$ and pick a suitable $A$. Here is a hint. If we tried $A = (1,2)$ then we would have $f(A)=(1,4)$. To make it a counterexample we should change $A$ a little, by adding some more stuff to $A$ while keeping $f(A)$ the same. So set $A = (1,2) \cup \text{[something]}$ where $f(A) = (1,4)$ still, meaning that the $[\text{something}]^2 \subset (1,4)$; and pick the $[\text{something}]$ to be a not open set.
I hope this is a good enough hint for (1). You can do something very similar for (3).
1 is not true. Consider $f(x) = x \sin(x)$ on $x \geq 0$, and $0$ on $x < 0$. Then $f([0,\infty)) = \mathbb{R}$.
The same example works for 3 if you use $(0, \infty)$ instead.
How to come up with this? I immediately thought "this has to be false", for reasons I'm not clear about; so I next tried to find the simplest possible answer. The simplest way would be if $f(A) = \emptyset$ but $A$ were nonempty - impossible. So the next simplest is if $f(A) = \mathbb{R}$ but $A$ is not open (and not closed, for question 3). We need $A$ not to be compact, because the continuous image of a closed bounded set is closed and bounded; the simplest non-compact nonopen sets are $[a, b)$.
Easy: just make $f$ take all values on $[0, \infty)$. How can this be done? Uurgh, this is annoying, I should just be able to draw a picture; but if you really must, something which oscillates ever wider as $x$ gets bigger. $\sin$ does the trick if we blow it up a bit.
More counterexamples:
For (1):
Take $f(x) = \left\lvert x\right\rvert$, $\forall x \in \mathbb{R}$ and $A = \left\{-\frac{3}{2}\right\}\cup\langle1, 2\rangle$. Then $f(A)=\langle1, 2\rangle$, which is open, but $A$ is not open.
For (3):
Take $f(x) = 1$, $\forall x \in \mathbb{R}$ and $A = \langle 0, 1\rangle$. Then $f(A) = \{1\}$, which is closed, but $A$ is not closed.
Note that (1) and (3) are in fact true when $f$ is an injective continuous function: if $f(A)$ is open then $A = f^{-1}\left(f(A)\right)$ which is open as a preimage of an open set. Same when $f(A)$ is closed.