The probability that you roll 4+ is the probability that you roll a 4, a 5, or a 6. Each of these events has probability $\frac{1}{6}$; so, the probability of rolling 4+ is $\frac{3}{6}=\frac{1}{2}$.
For your other scenario, note that the event that you get a 5+ when you are allowed on re-roll can be broken up in to two disjoint events: event $A$, in which you roll a 5+ on the first try; and event $B$, in which you roll a number 1-4 on your first attempt, and either a 5 or 6 on the second.
Then $P(A)=\frac{2}{6}=\frac{1}{3}$, since this is the event that you roll either a 5 or a 6. For $B$, we have
$$
P(B)=\frac{4}{6}\cdot\frac{2}{6}=\frac{2}{9},
$$
since you must roll a 1, 2, 3, or 4 on the first attempt and either a 5 or 6 on the second. So, overall, the probability of getting 5+ when you allow one re-roll is
$$
P(A\text{ or }B)=P(A)+P(B)=\frac{1}{3}+\frac{2}{9}=\frac{5}{9}.
$$
(Note that we have used here that $A$ and $B$ are disjoint possibilities.)
So, you are more likely to get a 5+ with a re-roll allowed than to get a 4+ with no re-roll.
Concerning a "closed" ( = finite summation) formula, start from
$$
\eqalign{
& N_b (s - n,m - 1,n) = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{
{\rm 1} \le {\rm integer}\;x_{\,j} \le m \hfill \cr
x_{\,1} + x_{\,2} + \; \cdots \; + x_{\,n} = s \hfill \cr} \right. = \cr
& = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{
0 \le {\rm integer}\;y_{\,j} \le m - 1 \hfill \cr
y_{\,1} + y_{\,2} + \; \cdots \; + y_{\,n} = s - n \hfill \cr} \right. \cr}
$$
where $N_b$ is given by
$$
N_b (s,r,m)\quad \left| {\;0 \leqslant \text{integers }s,m,r} \right.\quad =
\sum\limits_{\left( {0\, \leqslant } \right)\,\,k\,\,\left( { \leqslant \,\frac{s}{r+1}\, \leqslant \,m} \right)}
{\left( { - 1} \right)^k \binom{m}{k}
\binom
{ s + m - 1 - k\left( {r + 1} \right) }
{ s - k\left( {r + 1} \right)}\ }
$$
as explained in this related post.
Then the number of ways to obtain a sum $x \le s$ is given by
$$
\eqalign{
& M(x,m,n) = \sum\limits_{x\, \le \,s\,\left( { \le \,m\,n} \right)\,} {N_b (s - n,m - 1,n)} = \cr
& = m^{\,n} - \sum\limits_{0\, \le \,s\, \le \,x - 1\,} {N_b (s - n,m - 1,n)} = \cr
& = m^{\,n} - \sum\limits_{0\, \le \,s\, \le \,x - n - 1\,} {N_b (s,m - 1,n)} = \cr
& = m^{\,n} - \sum\limits_{0\, \le \,s\, \le \,x - n - 1\,} {\sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{s \over m}\, \le \,n} \right)}
{\left( { - 1} \right)^k \left( \matrix{ n \hfill \cr
k \hfill \cr} \right)\left( \matrix{
s + n - 1 - k\,m \cr
s - k\,m \cr} \right)} } = \cr
& = m^{\,n} - \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{s \over m}\, \le \,n} \right)} {\left( { - 1} \right)^k \left( \matrix{
n \hfill \cr
k \hfill \cr} \right)\left( \matrix{
x - 1 - k\,m \cr
x - n - 1 - k\,m \cr} \right)} \cr}
$$
and in fact $M(90,20,5)=3003$.
Note that, as explained in the cited related post, the problem has the geometric equivalent of finding:
the number of integral points on the diagonal plane $y_1, \cdots, y_n=s-n$, intercepted by a $n$-D cube
with side $[0,m-1]$.
The formula for $N_b$ corresponds to calculating the points on the whole plane ($k=0$) and subtracting those
pertaining to the surrounding cubes.
The geometric analogy clearly shows that $N_b(nm-s,m,n)=N_b(s,m,n)$.
Best Answer
As I see it, since there are $3$ equally likely choices for ranges $(1)$ and $(2)$, and $(6)$ equally likely choices for range $(3)$, you have $(3 \times 3 \times 6) = 54$ various combinations to examine. The challenge, as you noted in your posting is to avoid brute force, and find an approach that generalizes well.
I think that Stars and Bars kicks in here.
Let $x_1, x_2, x_3$ denote the numbers assigned to the 1st, 2nd, and 3rd ranges respectively.
You are trying to determine the probability that the randomly chosen $x_1 + x_2 + x_3$ will have a sum $> 88$.
The first thing to do, in accordance with the Stars and Bars format is to normalize the variables, so that each variable has a lower bound of $0$.
To that end, let:
$y_1 = x_1 - 30 \implies 0 \leq y_1 \leq 2.$
$y_2 = x_1 - 11 \implies 0 \leq y_2 \leq 2.$
$y_3 = x_1 - 42 \implies 0 \leq y_3 \leq 5.$
Then, $x_1 + x_2 + x_3 > 88 \iff y_1 + y_2 + y_3 > 5.$
Here, I would attack the problem by letting
$S$ = the number of possible combinations of terms $(54)$.
$T$ = the number of possible combinations such that
$y_1 + y_2 + y_3 \leq 5.$
Then, the desired ratio will be $\frac{S - T}{S}.$
Therefore, the problem has been reduced to enumerating $T$.
Introduce a new variable $(c)$ that is required to be a non-negative integer. Then, $T$ may be enumerated by enumerating the number of solutions to
$$y_1 + y_2 + y_3 + c = 5 ~~: ~~y_1, y_2 \leq 2, ~y_3 \leq 5.\tag1$$
Here, you have to use Inclusion-Exclusion.
First of all, the number of solutions to $a_1 + a_2 + \cdots + a_k = n$, where $a_1, \cdots, a_k$ can be any non-negative integers is given by $\binom{n + [k-1]}{k-1}.$
So the number of solutions to (1) above, where all of the upper bound constraints are ignored is
$$T_0 = \binom{8}{3} = 56.$$
The next thing to do, in accordance with Inclusion-Exclusion is to consider how many solutions there are when the upper bound of the 1st variable is violated, the upper bound of the 2nd variable is violated, and the upper bound of the 3rd variable is violated.
I will use the syntax that an equation bijects to another equation to indicate that they each have the same number of solutions.
The equation $y_1 + y_2 + y_3 + c = 5 ~: ~y_1 \geq 3$ bijects to $y_1 + y_2 + y_3 + c = 2 ~: ~y_1 \geq 0.$
This will have $\binom{2 + [4-1]}{[4-1]} = \binom{5}{3} = 10$ solutions. Since the upper bound on $y_2$ is identical to that on $y_1$, you have that altering $y_2$ (alone) will also have $10$ solutions. Note that alteration of $y_3$ is rejected here, because raising $y_3$ from a lower value of $(0)$ to a lower value of $(6)$ will not permit the sum $y_1 + y_2 + y_3 + c = 5$ to have any solutions.
This means that $T_1 = 2 \times 10 = 20$.
Therefore, the running total, with respect to Inclusion-Exclusion, at this point is
$$T_0 - T_1 = 56 - 20 = 36.$$
Now, in accordance with Inclusion Exclusion, you have to calculate $T_2$, which will represent the number of solutions where the upper bound of $2$ of the $3$ variables is violated.
However, if $y_1, y_2$ are each increased to a lower bound of $3$, then the equation $y_1 + y_2 + y_3 + c = 5$ will not have any solutions. Therefore, $T_2 = 0$.
Therefore, $$T = T_0 - T_1 = 36.$$
Therefore, the probability that the sum of the original $(3)$ numbers will be $> 88$ will be
$$\frac{54 - 36}{54} = \frac{1}{3}.$$
Note
To the best of my knowledge, the only two viable approaches to the problem as stated are [Stars and Bars + Inclusion-Exclusion] which I used or [generating functions] which I don't know anything about.
If you have never used Inclusion-Exclusion and never used Stars and Bars before, then you will need to study the pertinent articles (and perhaps other articles that you find by googling on "Inclusion-Exclusion" or "Stars and Bars Math"). This way you will stretch your intuition, which I regard as mandatory, for such an assignment.