Intuition/How to determine if onto or 1-1, given composition of g and f is identity. [GChart 3e P239 9.72] 
9.72. $A,B$ are nonempty sets. $f: A \rightarrow B$ and $g: B \rightarrow A$ are functions.
  Suppose $g \circ f = $ the identity function on $A$. (♦) Are the following true or false? 

$1.$ $f$ has a left inverse (ie: $g \circ f = id_A)$   $\iff$ $f$ is injective.
$2.$ $g$ has a right inverse  $\iff$ $g$ is surjective. (Question on its proof)
$3.$ $f$ need not be onto.
$4.$ $g$ need not be one-to-one.
$5.$ $f$ is onto $\iff$  $g$ is one-to-one.   
How would I determine truth or untruth for each, before proving or finding a counterexample? Moreover, what are the intuitions? I tried sketching possibilities from the given info, but it became desultory.   I'm not asking about formal arguments. 
Sources: Chartrand 3rd Ed P239 9.72 = 2nd Ed 9.48 and D Velleman P248 Thm 5.3.3 
 A: In terms of intuitions, I think that the best way to think about an injection is that it is a function that preserves difference; if two elements are different before having $f$ applied to them, then they are still different after having $f$ applied to them.  Symbolically:
$$x_1\neq x_2 \Rightarrow f(x_1) \neq f(x_2).$$  
Another way to think about this is as information: Knowing that $x_1\neq x_2$ is a sort of information.  If $f(x_1)=f(x_2)$, then observing a value $y\in \textrm{Range}(f)$ does not tell you which $x$ was input into $f$ to give the observed $y$.  Thus an injection can be said to preserve information.  A non-injective function collapses values in the domain.  
A typical example of a function that collapses values is a projection such as $f:\mathbb{R}^{2}\to\mathbb{R}$ given by $f(x,y)=x$.  The observation of $f(x,y)$ gives you some information about $\langle x,y\rangle$ - namely it tells you the value of the first coordinate.  But all information about the second coordinate has been lost.  
The way to think of a left inverse is that it undoes what the original function did.  So, $g$ is the left inverse of $f$ if $g\circ f$ is the same thing as doing nothing. In other words: $$g\circ f = \textrm{id}_{\textrm{dom}(f)}.$$
It should be apparent that a function $g$ can only undo the action of $f$ if the action of $f$ has not lost any information.  If information has been lost when calculating $f(x)$, then there is no way that $g$ can "know" where to return a value $y=f(x)$.  In other words, a function can only have a left inverse if it is injective. 
More formally, the application of a function cannot create information. As long as $ f $ is known, the value of $ f (x) $ cannot contain more information than value of $ x $ since $ f (x) $ can always be calculated from $ x $.  Once information has been lost, repeated application of other functions cannot recover it.  Thus, since the identity function obviously preserves information, the composition $ g\circ f $ must also preserve information, meaning that neither $ f $ nor $ g $ can lose information. The is an important caveat here though; it is not necessary that $ g $ never collapse values, only that the restriction of $ g $ to the range of $ f $ preserve information. What $ g $ does outside the range of $ f $ is irrelevant to $ g\circ f $. 
The converse of the above statement, that every injective function has at least one left inverse, is similar. You can think about this in terms of fibers.  If $ f :A\to B $ is a function, then for any $ b\in B $, the fiber of  $ f $ at $ b $ is the set $ \{a\in A \vert f (a)=b\} $. A function is injective of all the fibres have at most one element.  So, to construct a right inverse, just let $ g (b) $ be the unique element of the fibre of $ f $ at $ b$, if that fiber is nonempty (and any element of $ A $ you like if the fiber is empty.)
(If you want to understand why these sets are called fibers, consider the fibers of the projection in my example above.)
The dual notions of surjections and right inverses are a little less easy to intuit in this way, but can be made sense of using fibers again. 
If $g:B\to A$ is known to be a surjection, then every fiber $\{b\in B\vert g(b)=a\}$ is non empty.  To construct a right inverse of $g$, all you need to do is find an $f$ that takes each $a\in A$ to some element of the corresponding fiber.  As long as you accept the Axiom of Choice, this can always be done, so every surjection has a right inverse. 
Conversely, it's pretty easy to see that any function with a right inverse must be a surjection.  The range of $g\circ f$ is necessarily a subset of the range of $g$, but if $g\circ f=\textrm{id}_{\textrm{dom}(f)}$, then the range of $g\circ f$ must be the range of $\textrm{id}_{\textrm{dom}(f)}$, which is the entire domain of $f$.  Thus the range of $g$ must be the whole domain of $f$, and $g$ must be a surjection. 

So, where does this leave your questions?
(1) and (2) are a little odd.  We know from the statement of that question that $g\circ f=\textrm{id}_{A}$, and so $f$ has a left inverse ($g$), and $g$ has a right inverse ($f$). Thus in both cases, the left hand side of the biconditional is assumed true, and so the biconditionals are true just in the cases that the right hand sides are true. 
However, since the things I have written above proves the biconditionals directly, there's noting much to add.  
(3) and (4) are more interesting.  We know that $f$ preserves difference/information.  But need it be surjective?  Well, there's no reason to think it must.  The fact that there may be points in the codomain of $f$ (i.e. $B$) which do not lie in the range of $f$ seems irrelevant to the question of information.  
Consider the function $f:\mathbb{R}\to\mathbb{R}^{2}$ given by $f(x)=\langle x,x\rangle$.  If I told you that $f(x)=\langle 3,3\rangle$, you would have no trouble telling me the value of $x$.  And if told you that $f(x)=\langle 3,4\rangle$, while you could not tell me the value of $x$, you could tell me that I was wrong - that $\langle 3,4\rangle$ could not possibly the value of $f(x)$.  So, you are able to construct a left inverse to $f$, even though $f$ is not surjecive. 
I'll leave the rest for you, unless you have more questions. 
A: How to determine truth values? By finding a proof, or a counterexample.
How does one do that? How does one approach a problem in mathematics to begin with? One finds some examples, plays with them, tests whether or not they can be modified to be counterexamples (in the case they are not counterexamples to begin with), and if one suspects that the answer is no, one sits down to write the proof.
Whence one's proof cannot proceed, one constructs a better counterexample, or a better argument.
Repeat if necessary.

In this case, pick some $A$ and $B$ and some $f$ and $g$, try to vary the parameters of what are the sizes of $A$ and $B$ and what do $f$ and $g$ do. See if you can find a counterexample to each of these statements, if not sit to write a careful proof.
Go over the arguments, and when you got stuck, analyze why that happened. Maybe it was because you needed one of the functions to have an extra property, maybe you needed one of the sets to have an extra property. From these generate a counterexample. If your counterexample doesn't work, examine carefully why that is not a counterexample, and if need be, restart your proof using the new insights that you have gained.
