I assume it should be true (and known) that $a + ab = a$.
Assuming this holds, let $x = a+(b+c)$ and $y = (a+b)+c$. We want to show that $x = y$, and following the hint we reduce to showing $x = xy = y$.
I claim that $ax = a, \ bx = b, \ cx = c$, and likewise for $y$. We check for $ax$:
$$ ax = aa + a(b+c) = a + a(b+c) = a$$
Likewise, for $bx$:
$$ bx = ba + b(b+c) = ba + (bb+bc) = ba + (b+bc) = ba + b = b$$
The remaining checks are analogous.
Using these identities, you can derive that anything made up of $a,b,c,+,.$ does not change when multiplied by $x$, in particular $yx = x$:
$$ yx = ((a+b)+c)x = (a+b)x+cx = (ax+bx)+cx = (a+b)+c = y$$
You can use a symmetric argument to conclude that $yx = xy = x$, and hence the claim follows.
For products, you can use a similar trick. Let $x = a.(b.c)$ and $y = (a.b).c$. I claim that $ x = x + y = y$. To see this, first note that $x + a = a$ (because $x+a = a+ a.(...) = a$. Secondly, $x+b = b$, because
$$ x+b = a.(b.c) + b = a.(b.c) + a.b + a'.b = a.(b.c+b) + a'.b = a.b + a'.b = b$$
(I hope this is legit). Likewise, $x+c = c$. Finally, $x + y = y$, because:
$$ y = (a.b).c = ((a+x).(b+x)).(c+x) = (a.b + x).(c+x) = (a.b).c + x = y + x$$
(I used the identity $(u+t).(v+t) = u.v + t.u + t.v + t.t = u.v +t$). The proof that $y+x = x$ is symmetric.
Although, yes, in practice it doesn't usually matter what order you write a truth table, this author seems to tacitly be making the assumption that you must write a truth table in Lexicographical Order. I.e, flip the furthest bit first.
Thought of in a different way, if you interpret the entries as a binary number, they are written from smallest to largest. $000_2 = 0, ~001_2=1, ~010_2=2, ~011_2=3,~\dots$
By making sure such truth tables are written in a very specific order, the representation the author refers to is then well-defined and unique.
Thinking of a boolean function on three variables as a function from $\mathbb{F}_2^3\to \mathbb{F}_2$, their notation $F=\sum(1,2,4,7)$ is just saying that $F(0,0,0)=0,~F(0,0,1)=1,~F(0,1,0)=1,~F(0,1,1)=0,~F(1,0,0)=1,~F(1,0,1)=0,~F(1,1,0)=0,~F(1,1,1)=1$ by simply only listing the ones with an output of one in shorthand notation based on where they appeared on the list. ($F(1,1,1)$ for example being the seventh entry on the list, where we start counting from zero, hence the $7$ in $\sum(1,2,4,7)$)
Best Answer
Let's call your expression $f$, $$ \DeclareMathOperator{\and}{~And~} \DeclareMathOperator{\or}{~Or~} \DeclareMathOperator{\nt}{~Not~} \DeclareMathOperator{\T}{{\color{blue}T}} \DeclareMathOperator{\F}{{\color{red}F}} f = \bigg((A \and \nt C) \or \nt (A \and \nt C)\bigg) \and \nt (A \and \nt B) $$
$$\begin{array} {c|ccc|c} & A & B & C & f \\ \hline \text{Row 1} & \T & \T & \T & \T \\ \text{Row 2} & \T & \T & \F & \T \\ \text{Row 3} & \T & \F & \T & \F \\ \text{Row 4} & \T & \F & \F & \F \\ \text{Row 5} & \F & \T & \T & \T \\ \text{Row 6} & \F & \T & \F & \T \\ \text{Row 7} & \F & \F & \T & \T \\ \text{Row 8} & \F & \F & \F & \T \\ \end{array}$$
Notice how your sum of minterms has 6 terms. Your truth table also has 6 "true" entries. To construct minterms, you just list the "true" cases:
$$\begin{align}f &= (\text{Row 1}) \or (\text{Row 2}) \or (\text{Row 5}) \or (\text{Row 6}) \or (\text{Row 7}) \or (\text{Row 8})\\ &= ABC \or AB\overline C \or \overline ABC \or \overline AB\overline C \or \overline A\overline BC \or \overline A\overline B\overline C\end{align}$$
To construct the product of maxterms, look at the value of $\nt f$. There are 2 cases when $f$ is false:
$$\begin{align}\nt f &= (\text{Row 3}) \or (\text{Row 4}) \\ &= A\overline B C \or A \overline B \overline C\end{align}$$
Now apply DeMorgan's: $$\begin{align} \nt\nt f &= \nt(A\overline B C \or A \overline B \overline C) \\ f &= (\overline A\or B\or \overline C) \and (\overline A \or B \or C) \end{align}$$
Your given answer for "product of maxterms" is the value of $f(\nt A, \nt B, \nt C)$. Perhaps you are working with a different definition.