For the first part:
$$100a+10b+c = a+b^2+c^3$$
implies that $$99a + b(10-b)=(c-1)c(c+1)$$
Now, $b(10-b)\leq 25$, with $b(10-b)=0,9,16,21,24,25$. So you want $c$ so that $$(c-1)c(c+1)\equiv 0,9,16,21,24,25\pmod{99}.$$
Now, $(c-1)c(c+1)$ is divisible by $3$, as is $99$, so you really need $$c^3-c\equiv 0,9,21,24\pmod {99}, \,c^3-c\geq 100$$.
This can be done by trial and error.
We know $c>4$. (If $c\leq 4$ then we get $a=0$, which you might want to include.)
$c=5$ gives $c^3-c\equiv 21\pmod {99}$.
So $\overline{abc}=135, 175$ are solutions.
$6^3-6\equiv 12\pmod{99}$, so $c\neq 9$.
$7^3-7\equiv 39\pmod{99}$, so $c\neq 7$.
$8^3-8\equiv 9\pmod{99}$ so $(a,b,c)=(5,1,8)$ and $(a,b,c)=(5,9,8)$ are solutions.
$9^3-9\equiv 27\pmod{99}$. So $c\neq 9$.
This yields the four solutions: $135, 175, 518, 598$.
(Allowing $a=0$, we get additional answers $\overline{abc}=000,001,043,063$.)
A start on the bonus question.
For $n$ digits, the largest sum $a_1+a_2^2+\cdots a_2^n$ is $9+9^2+\cdots +9^n = \frac{9}{8}(9^n-1)$. The smallest $n$-digit number is $10^{n-1}$.
So if $10^{n-1}>\frac{9}{8}(9^n-1)$, you can be sure there is no solution. But $$\frac{9}{8}(9^n-1)< \frac{9^{n+1}}{8},$$ so if $10^{n-1}>9^{n+1}/8,$ there is no solution. This is true if $(n-1)\log_9(10)>(n+1)-\log_9 8$ or $$\log_9(10) > 1+\frac{2-\log_9 8}{n-1}\\\log_9(10/9)>\frac {2-\log_9 8}{n-1}$$ or $$n>1+(2-\log_9(8))\log_{10/9}(9)\approx 22.97$$
So there are no solutions with more than 22 digits. That limits it to a finite, if large, problem.
Indeed the answer I believe is 50268 as @Hw Chu says. The reason is as follows. Note that having 5 in the tens place doesn't give the students enough information to actually deduce the actual number (and one must know all the numbers to say with certainty it is divisible by 59). So after the teacher asks if anyone knows a prime factor and 3 people raise their hands, we have the following configuration that is known to all:
5_ _ _ [even]
Now the teacher asks if anyone knows a composite factor. The only two people that can answer and the ones that do are Dick and Carl. It is pretty clear why this is. Eason has only access to the last digit (which is not enough to answer the question, Bob has access to the thousands place and hundreds which is not enough and for similar reasons Alice cannot answer as well). Now, this means that Carl, with his knowledge of the hundreds and tens place and knowing that the last digit is even, is able to deduce with certainty that the number is divisible by 4.
So the possibilities of the last digit are as follows: 0, 2, 4, 6, 8. Now Carl must be able to reduce the possibilities with his knowledge so that no matter what the number is, the last two digit as a number itself is divisible by 4. Hence, he must be able to reduce the 5 to 3 choices. This is obvious as 4 is not enough (if this isn't clear, it will be clear in a second). Now to do so, Carl must see 2 even numbers. So let us for example say he sees 4, 6. Then in particular, he leaves 0 and 2 as possible choices for the last number. But if _0 is divisible by 4, then _2 is not. So this cannot be the case. In fact, the 3 possibilities he leaves as must be an arithmetic progression differing by 4. Hence, it must be 0, 4, 8. That is, he sees 2, 6. Now this doesn't let us know in what particular order he sees them.
But now Bob knows Carl sees {2, 6} as Bob can deduce this himself from seeing Carl raise his hand. So Bob must be able to deduce exactly what the number is from the information remaining. Carl knows the order in which the numbers appear {2, 6} but Bob knows everything Carl knows because he knows the numbers Carl sees AND knows the order by seeing the hundreds place. So let's think about how Bob can deduce the final number. He can see that the last digit must be 0, 4, or 8. But he also knows Carl cannot deduce the number is divisible by 8. Ah! But Eason also didn't raise his hand for 2 composite numbers so the last digit CANNOT be 0. So we can deduce that Bob can deduce the last 3 digits are from the following: 624, 628, 264, 268. However, since he knows the exact number, i.e. knows exactly what the last 3 numbers are, he must also see 2 even numbers to limit the choice to one possibility. But to us, he limits it to the following: 50624, 50628, 50264, 50268. Only one of these is divisible by 59, which is 50268.
Best Answer
We want to minimise $\cfrac {100a+10b+c}{a+b+c}$
This is equal to $$1+\frac{99a+9b}{a+b+c} $$ and this is clearly least when c is greatest. So we have $c=9$.
We then rewrite the fraction as $$100-\frac{90b+99c}{a+b+c}$$ which is least when $a$ is least. So we have $a=1$.
Isolating $b$ as above gives $$10+\frac {90a-9c}{a+b+c}$$ Since $a$ is non-zero the numerator is positive and the minimum is obtained when $b=9$.