I would say that it's almost a sham - not in the sense that it's worthless, just that it markets to people who are apt to exaggerate its worth. Let's consider a few of the things that people might do with such 'bought' primes: they might try to decrypt something, encipher something, or have some sort of sentimental value for a particular prime.
First: trying to decrypt something by buying lists of 400 digit primes. It seems to me that somehow, we have figured out that a message is enciphered using a 400 digit prime (likely the product of that prime and another, larger prime). We are trying to brute force attack it - but how many 400 digit primes are there? By the prime number theorem, we know there are about $\frac{x}{\log x} $ primes up to x, so we consider $\dfrac{10^{400}}{400 \log 10} - \dfrac{10^{399}}{399 \log 10}$, which is of the order $10^{398}$. Even if the list had each of these primes on it, just trying them all out is computationally improbable (to be... generous). It's akin to saying: we know this door is locked by a key of about this size, so let's get every such key in the world and try them.
Secondly: we are trying to encipher something. I will assume that we are going to use some public key cryptography and are trying to come up with something incoveniently large to decrypt, perhaps using RSA. But then we will probably use 2 primes, resulting in a number of over 800 digits. That's annoyingly large. Worse, it is very easy to find your own 400 digit prime. One would expect to find one at random, by checking consecutive odd 400 digit numbers, every thousandth guess or so ($1/{400 \log 10}$ numbers on average need to be checked, not counting the consideration of only even numbers). One would assume that needing such incredible security would mean that you at least own a computer, and therefore have the capabilities of finding your own large primes.
Thirdly, sentimental value. This reminds me of the 'Name a Star after Someone' campaign, like here. The author John Allen Paulos, I think, once quipped in his book Innumeracy that he should like to offer to name numbers after people, maybe charging more for prime numbers, etc.
But in regards to your fear of buying a number that someone else has bought - suppose that they choose 100 different 400 digit primes randomly for each list they sell. Then it wouldn't be until something like their $10^{200}$th sale that we would expect someone to have bought the same number as you. However, if we consider this instead like the same-birthday 'paradox,' this number would be several dozen orders of magnitude reduced, and yet still meaninglessly large. There's isn't enough money in the world to get to that level.
In fact, they could name a different 400 digit prime after each human, every day, for longer than our sun will be around. I should start selling primes on ebay.
But the awards for computing larger and larger primes are not because of the use of the resulting prime number, but instead because it inspires development of computational and algorithmic methods. Finding large primes quickly becomes very, very challenging. I believe the current awards is for a prime of 100,000,000 digits, a very challenging number even to store in a computer's memory, let alone manipulate.
Consider the following (mod 10)
$$0^2=0,1^2=1,2^2=4,3^2=9,4^2=6,5^2=5,6^2=6,7^2=9,8^2=4,9^2=1$$
Since the tens digit obviously doesn't matter for what the one digit is, these are all the ways that a square can end.
For the question about an even tens digit, just square $(\sum a_i 10^i)^2=a_0^2+2a_0a_1\cdot 10+\cdots$ You have an even number in front of the $10$, which switches to odd when $a_0^2$ has an odd tens digit. For the question about ending in a $5$ causing the $10$'s digit to be a $2$, notice that if $a_0=5$, then the first term is $25$ and the second is $100a_1$ which doesn't effect the tens spot.
Best Answer
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.