The theorems below are from Peter Smith's a Gentle Intro to Category theory, but I am not sure why the proof are like this. (I understand that most of my questions may appear unsophisticated, that's because I do not have any maths background, so please bear with me...)
Specifically:
In Thm 70, I understand that by def. of a terminal object there is only arrow $A\times B \to 1$, but why then is there exactly one arrow $A\to 1^B$? It seems that it has something to do with the bijection/one-one correlation between $A\to 1^B$ and $A\times B \to 1$ - I get that this implies the two are thus isomorphic, but I am not sure how this comes into play here.
As far as I understand, isomorphic objects are supposed to preserve the same kind of relations within their respective members, and are 'structurally speaking identical', but still this doesn't seem to explain it.
In Thm 69, I am not sure why the injectivity/monic and surjectivity/epic are proved the way Smith did. As far as I understand, an arrow $f: a\to b$ is monic iff $f\circ g=f\circ h$ implies $g=h$, and epic iff for any pair of arrow $g,h:b\to c$, $g\circ f=h\circ f$ implies $g=h$.
So in the monic proof, $\bar{g}=\bar{h}$ is analogues to assuming $f\circ g =f\circ h$ in a normal monic proof? But why? Say we call the map $f: g\mapsto \bar{g}$, shouldn't we assume $f\circ g =f\circ h$ (not sure what $g$ and $h$ would be though...), then trying to prove $g=h$?
Likewise, the way epic is proved differs from how I thought it should be proved (assume $g\circ f=h\circ f$, where $f:g \mapsto \bar{g}$, then prove that $g=h$).
Any help would be greatly appreciated, thank you so much!
Theorem 70: If $\mathscr C$ is a Cartesian closed category with terminal object $1$, then for all $A, B, C\in \mathscr C$, $1^B \cong 1$.
By Theorem 69, for each $A$, there is a one-one correlation between arrows $A\to 1^B$ and arrows $A\times B \to 1$. But since $1$ is terminal, there is exactly one arrow $A\times B \to 1$; hence for each $A$, there is exactly one arrow $A\to 1^B$. Therefore $1^B$ is terminal, and hence $1^B\cong 1$.
Theorem 69: If there exists an exponential of $C$ by $B$ in the category $C$ , then, for any object $A$ in the category, there is a one-one correlation between arrows $A \times B \to C$ and arrows $A \to C^B$.
There is also a one-one correlation between arrows $A\to C^B$ and arrows $B\to C^A$.
Proof: By definition of the exponential $[C^B, ev]$, an arrow $g: A\times B \to C$ is associated with a unique 'transpose' $\bar{g}:|A \to C^B$ making the diagram commute.
The map $g\mapsto \bar{g}$ is injective. For suppose $\bar{g}=\bar{h}$. Then $g=ev \circ (\bar{g}\times 1_B)=ev\circ (\bar{h}\times 1_B)=h$.
The map $g\mapsto \bar{g}$ is also surjective. Take any $k: A\to C^B$, then if we put $g=ev\circ (k \times 1_B)$,$\bar{g}$ is the unique map such that $ev\circ (\bar{g}\times 1_B)=g$, so $k=\bar{g}$.
Hence $g\mapsto \bar{g}$ is the required bijection between arrows $A\times B\to C$ and arrows $A\to C^B$, giving us the first part of the theorem.
For the second part, we just note that arrows $A\times B \to C$ are in one-one correspondence with arrows $B\times A \to C$, in virtue of the isomorphism between $A\times B$ and $B\times A$. We then apply the first part of the theorem.
For your second question, you are confusing the arrows of the category being monic or epic with the bijection between the hom-sets. So let me be more specific about it, when he proves that the map $g\mapsto\overline{g}$ is injective, that does not means in any way that $\overline{g}$ is monic in the category $\mathcal{C}$. It means that the thing which associate $\overline g$ from $g$ is injective, or in other words, two different maps $g,h : A\times B \to C$ will give two different maps $\overline{g},\overline{h} : A\to C^B$. Read the proof again, and you will realise that it corresponds exactly to that. The exact same distinction goes for the surjection part of the proof
For your first question, you need to remember that "terminal object" exactly means that there is a unique arrow from ay object to this object, and that it is unique up to isomorphism. Thus when you have a terminal object $1$, if you want to prove that $1^B$ is isomorphic to $1$, you really just need to prove that $1^B$ is also a terminal object, that is for any object $A$, there is a unique morphism $A \to 1^B$. By definition of $1^B$ (or by theorem 69 if you will), the set of morphism $A \to 1^B$ is in bijection with the set of morphism $A\times B \to 1$. By definition of terminal object, there is exactly one morphism $A\times B \to 1$, and thus, by the bijection there is exactly one morphism $A \to 1^B$. By definition, this exactly means that $1^B$ is terminal, hence $1\simeq 1^B$