(a) We claim that $(a^2-1, a^2, a^2+1, a^2+2)$ works for $a = 2^n + 1$ where $n \ge 1$. Indeed, we can verify each number in the set:

1. $a^2 - 1 = (2^n + 1)^2 - 1 = 2^{2n} + 2^{n+1} + 1 - 1 = 2^{n+1} + (2^n)^2$
2. $a^2 = (2^n + 1)^2 = 2^{2n} + 2^{n+1} + 1 = 2^{n+2} + (2^n - 1)^2$
3. $a^2 + 1 = (2^n + 1)^2 + 1 = 2^{2n} + 2^{n+1} + 1 + 1 = 2^0 + a^2$
4. $a^2 + 2 = (2^n + 1)^2 + 2 = 2^{2n} + 2^{n+1} + 1 + 2 = 2^1 + a^2$

Thus, all four numbers are good, proving that there are infinitely many sets of 4 consecutive good numbers.

(b) We claim that the solutions are $(1, 2, 3, 4, 5)$, $(2, 3, 4, 5, 6)$, $(8, 9, 10, 11, 12)$, $(9, 10, 11, 12, 13)$, $(288, 289, 290, 291, 292)$, and $(289, 290, 291, 292, 293)$.

To prove this, we need to show that these sets are the only possible sets of 5 consecutive good numbers. We start by examining the function $f(n) = (2^n + 1)^2 + 3$ and show that it is good if and only if $n = 1$ or $n = 4$.

Proof:
1. For $n = 1$, $f(1) = (2^1 + 1)^2 + 3 = 3^2 + 3 = 9 + 3 = 12 = 2^2 + 2^2$, which is good.
2. For $n = 4$, $f(4) = (2^4 + 1)^2 + 3 = 17^2 + 3 = 289 + 3 = 292 = 2^8 + 6^2$, which is good.

Now, assume $n \ge 4$ and suppose $f(n)$ is good, i.e., $(2^n + 1)^2 + 3 = 2^x + y^2$. This implies:
$$y^2 = 2^{2n} + 2^{n+1} + 4 - 2^x$$

We split into cases based on the value of $x$:

Case 1: $x \le n + 1$
- Subcase 1.1: $x = 0$
$$y^2 = 2^{2n} + 2^{n+1} + 3 = (2^n + 1)^2 + 2$$
No two squares differ by 2, so no solutions here.

- Subcase 1.2: $x = 1$
$$y^2 = 2^{2n} + 2^{n+1} + 2 = (2^n + 1)^2 + 1$$
No two squares differ by 1, so no solutions here.

- Subcase 1.3: $2 \le x \le n + 1$
$$(2^n)^2 < y^2 < (2^n + 1)^2$$
This means there are no solutions here.

Case 2: $x \ge n + 2$
Let $x = n + 1 + \alpha$ where $\alpha \ge 1$. We see from the equation that $y$ is even, so let $y = 2z$. Furthermore, let $m = n - 1$. The equation reduces to:
$$z^2 = 2^{2m} - 2^m(2^\alpha - 1) + 1$$
This means that $z$ is odd and less than $2^m$, so let $z = 2^m - k$ where $k \ge 1$ is odd. Then,
$$z^2 = 2^{2m} - 2^m(2k) + k^2$$
$$2^m(2k - 2^\alpha + 1) = (k - 1)(k + 1)$$

We have some subcases:

- Subcase 2.1: $k \equiv 1 \pmod{4}$
$$2^m \mid (k - 1)(k + 1)$$
$$2^{m-1} \mid (k - 1)(\frac{k + 1}{2})$$
Since $(k + 1)/2$ is odd, we have $2^{m-1} \mid k - 1$, so write $k = 1 + 2^{m-1}\beta$, where $\beta \ge 0$. Plugging this into the equation gives:
$$2^m\beta - 2^\alpha + 3 = 2^{m-2}\beta^2 + \beta$$
The left side is odd, so the right side is odd. Since $m = n - 1 \ge 3$, $\beta$ must be odd.

- Subcase 2.1.1: $\beta = 1$
$$2^m - 2^\alpha + 3 = 2^{m-2} + 1$$
$$3(2^{m-2} + 1) = 2^\alpha + 1$$
This has the unique solution $m = 3$ and $\alpha = 3$. This gives $n = 4$, which is a solution.

- Subcase 2.1.2: $\beta = 3$
$$3 \cdot 2^m - 2^\alpha + 3 = 2^{m-2} \cdot 9 + 3$$
$$3 \cdot 2^{m-2} = 2^\alpha$$
This has no solutions.

- Subcase 2.1.3: $\beta \ge 5$
$$3 - 2^\alpha = 2^{m-2}(\beta^2 - 4\beta) + \beta \ge 5 \cdot 2^{m-2} \ge 10$$
This has no solutions.

- Subcase 2.2: $k \equiv 3 \pmod{4}$
$$2^m \mid (k - 1)(k + 1)$$
$$2^{m-1} \mid (k + 1)(\frac{k - 1}{2})$$
Since $(k - 1)/2$ is odd, we have $2^{m-1} \mid k + 1$, so write $k = -1 + 2^{m-1}\beta$, where $\beta \ge 1$. Plugging this into the equation gives:
$$2^m\beta - 2^\alpha - 1 = 2^{m-2}\beta^2 - \beta$$
The left side is odd, so the right side is odd. Since $m = n - 1 \ge 3$, $\beta$ must be odd.

- Subcase 2.2.1: $\beta = 1$
$$2^m - 2^\alpha - 1 = 2^{m-2} + 1$$
$$3 \cdot 2^{m-2} = 2^\alpha$$
This has no solutions.

- Subcase 2.2.2: $\beta = 3$
$$3 \cdot 2^m - 2^\alpha - 1 = 2^{m-2} \cdot 9 - 3$$
$$2^{m-2} \cdot 3 = 2^\alpha - 2$$
For the right side to be divisible by 3, we need $3 \mid \alpha$.

- Subcase 2.2.2.1: $\alpha = 3$
We get $m = 3$, so $n = 4$, which is a solution.

- Subcase 2.2.2.2: $\alpha \ge 6$
The right side is $2 \pmod{4}$, so we must have $m = 3$. But this is a contradiction to size, so no solutions here.

- Subcase 2.2.3: $\beta \ge 5$
$$-1 - 2^\alpha = 2^{m-2}(\beta^2 - 4\beta) - \beta \ge 2(\beta^2 - 4\beta) - \beta \ge 5$$
This has no solutions.

This completes all the cases, so the only solution for $n \ge 4$ is $n = 4$. This completes the proof of the claim. $\blacksquare$

Let's return to solving the original problem. We'll be needing two recurring lemmas.

Lemma: Suppose $n$ is good. Then, $n \equiv 2 \pmod{4}$ implies either $n = a^2 + 1$ for odd $a$, or $n = a^2 + 2$ for even $a$. Similarly, if $n \equiv 3 \pmod{4}$, then $n = a^2 + 2$ for odd $a$.

Proof: The key is that if $x \ge 2$, then $2^x + y \equiv 0, 1 \pmod{4}$. So if $n = 2^x + y \equiv 2, 3 \pmod{4}$, then $x = 0$ or $x = 1$. Working out the details of mod $4$ gives the result. $\blacksquare$

Lemma: Suppose $a^2$ is good where $a$ is odd. Then either $a = 1$, or $a = 2^n + 1$ for some $n \ge 1$.

Proof: Suppose $a^2 = 2^x + y^2$ where $x \ge 1$. Then, we have $a^2 = 2^x + y^2$, so
$$(a - y)(a + y) = 2^x$$
Now, $a - y \equiv a + y \pmod{2}$, and since $x \ge 1$, this means that they both must be even. Thus, $a - y = 2^p$ and $a + y = 2^q$ for some $p, q \ge 1$, so $a = \frac{2^p + 2^q}{2}$. Since $a$ is odd, this means $a = 2^n + 1$ for some $n \ge 1$.

Now, if $x = 1$, then $a^2 = 1 + y^2$. This means $y = 0$ and $a = 1$, as desired. This completes the proof of the lemma. $\blacksquare$

Suppose that $(a_1, a_2, a_3, a_4, a_5)$ are 5 consecutive positive integers that are all good. We have 4 cases.

Case 1: Suppose $a_1 \equiv 0 \pmod{4}$. We see that $a_4 \equiv 3 \pmod{4}$ is good, so $a_4 = a^2 + 2$ for some odd $a$. Thus, the numbers $(a^2 - 1, a^2, a^2 + 1, a^2 + 2, a^2 + 3)$ are all good. Thus, $a^2$ is good and $a$ odd, so $a = 1$ or $a = 2^n + 1$ for $n \ge 1$. We can't have $a = 1$ as $1^2 - 1 = 0$ is not good, so we must have $a = 2^n + 1$. Thus, $a^2 + 3 = (2^n + 1)^2 + 3$ is good, so by the claim, we have either $n = 1$ or $n = 4$. These give the two solutions:
$$\boxed{(8, 9, 10, 11, 12) \quad \text{and} \quad (288, 289, 290, 291, 292)}$$

Case 2: Suppose $a_1 \equiv 1 \pmod{4}$. We see that $a_3 \equiv 3 \pmod{4}$ is good, so $a_3 = a^2 + 2$ for some odd $a$. Thus, the numbers $(a^2, a^2 + 1, a^2 + 2, a^2 + 3, a^2 + 4)$ are all good. Thus, $a^2$ is good and $a$ odd, so $a = 1$ or $a = 2^n + 1$ for $n \ge 1$. If $a = 1$, then we get the solution:
$$\boxed{(1, 2, 3, 4, 5)}$$
Now suppose $a = 2^n + 1$ for $n \ge 1$. Thus, $a^2 + 3 = (2^n + 1)^2 + 3$ is good, so by the claim, we have either $n = 1$ or $n = 4$. These give the two solutions:
$$\boxed{(9, 10, 11, 12, 13) \quad \text{and} \quad (289, 290, 291, 292, 293)}$$

Case 3: Suppose $a_1 \equiv 2 \pmod{4}$. Then, $a_2 \equiv 3 \pmod{4}$ is good, so $a_2 = a^2 + 2$ for some odd $a$. Thus, the numbers $(a^2 + 1, a^2 + 2, a^2 + 3, a^2 + 4, a^2 + 5)$ are all good. Now, $a^2 + 5 \equiv 2 \pmod{4}$ is good, so by the first lemma, we have the following two subcases:

- Subcase 3.1: Suppose $a^2 + 5 = b^2 + 1$ for odd $b$. Then, $b^2 - a^2 = 4$, so $a = 0$, which isn't odd.

- Subcase 3.2: Suppose $a^2 + 5 = b^2 + 2$ for even $b$. Then, $b^2 - a^2 = 3$, so $a = 1$. This gives the solution:
$$\boxed{(2, 3, 4, 5, 6)}$$

Case 4: Suppose $a_1 \equiv 3 \pmod{4}$. Then, $a_1 \equiv 3 \pmod{4}$ is good, so $a_1 = a^2 + 2$ for some odd $a$. Thus, the numbers $(a^2 + 2, a^2 + 3, a^2 + 4, a^2 + 5, a^2 + 6)$ are all good. We see that $a^2 + 6 \equiv 3 \pmod{4}$ is good, so $a^2 + 6 = b^2 + 2$ for odd $b$. Thus, $b^2 - a^2 = 4$, so $a = 0$, which isn't odd. So there are no solutions here.

We've shown that all the solutions must be in the claimed list, so we're done.