(32.) Let $H \le$ group G and let $a, b \in G.$ Prove or disprove.
If ${aH= bH},$ then $Ha^{-1} = Hb^{-1}.$$\color{blue}{Ha^{−1}} = \{\color{magenta}ha^{−1} | h ∈ H\} = \{\color{magenta}{h^{−1}}a^{−1} | h ∈ H\} = \{(ah)^{−1} | h ∈ H\} = \color{blue}{(aH)^{-1}}$
Just replace $a$ with $b$ overhead to induce $\color{blue}{Hb^{−1} = (bH)^{-1}}$.Hence $aH= bH \iff (aH)^{-1} =(bH)^{-1} \color{blue}{\iff Ha^{-1} = Hb^{-1}} $
(1.) The trick or hinge looks like $\color{magenta}{h \in H \iff h^{-1} \in H}$. How do you preordain this?
(35.) Show that |left cosets| = |right cosets| of a subgroup H of a group G.
From the proof in Exercise 32, $\color{blue}{Ha^{−1} = (aH)^{-1}}$. This shows $φ(aH) = Ha^{−1}$, the map φ of the collection of left cosets into the collection of right cosets, is well defined.
(2). Where does the bijection isomorphism $φ(aH) = Ha^{−1}$ loom from?
I know how to prove 1-1 and onto, hence I omit this. I'm not querying the proof.
(3.) What's the intuition for (35.)? How does it relate to (32.)? Are there pictures?
For 32):
Observe that the hypothesis $aH=bH$ implies $a\in bH$. Then $a=bh_0$ for some $h_0\in H$. So $$\color{brown}{a^{-1}=h_0^{-1}b^{-1}}.\qquad (*)$$
Now if $x\in Ha^{-1}$ then $x=h_1a^{-1}$ for some $h_1\in H$. But substituting $(*)$ we will get $x=\color{brown}{h_1h_0^{-1}b^{-1}}$, which clearly gives $x\in Hb^{-1}$. Hence $Ha^{-1}\subseteq Hb^{-1}$.
Similarly one can show $Ha^{-1}\supseteq Hb^{-1}$.
For 35):
Define $ah\longmapsto bh,$ to map $aH\to bH$, now you can prove this is a bijection. Similar for $Ha\to bH$ by mapping $ha\longmapsto bh$. Same for any other combination.
Remark:
You can't say that $\varphi$ could be an isomorphism because $aH$ is not a group, save at $a=e$.
Commenting:
I saw your new solutions I think that they are right. However:
For (1.), what your have done is equivalent to what I did, but I have used a more primitive approach which says that between two sets $N,M$, they are $N=M$ if and only if $N\subseteq M$ and $N\supseteq M$.
For (2.), I explain details:
Injectivity: If we name the map used as $f(ah)=bh$ and supposing $f(ah_1)=f(ah_2)$ then $bh_1=bh_2$. So by left's multiplying with $a^{-1}$ we have $h_1=h_2$. So we have $ah_1=ah_2$.
Surjectivity: Taking $bh_3\in bH$ we have $ah_3\in aH$ such that $f(ah_3)=bh_3$.
For (3.), I would choose a good example to get acquainted, such as $S_3$, the group of six permutations on three labels. The group $S_3$ is the smallest non commutative group where talking about non trivial properties of cosets can be neatly illustrated.
If these ideas still are not enough, let me know.