a. For $(m, n) = (6, 10)$, $(12, 15)$ and $(30, 78)$, $p$ is equal to
$$\frac{10^2 + 6^2}{8} = 17, \quad \frac{15^2 + 12^2}{9} = 41 \text{ and } \frac{78^2 + 30^2}{72} = 97.$$
respectively.

b. Let $k = \sqrt{n^2 - m^2}$. Then $(k, m, n)$ is a Pythagorean triple. Therefore, there exist positive integers $d, x$ and $y$ such that $(x, y) = 1$, $n = d(x^2 + y^2)$ and $m = d(x^2 - y^2)$ or $m = 2dxy$.

If $m = d(x^2 - y^2)$, then $p = \frac{d(x^4 + y^4)}{xy}$. Since $(x, y) = 1$, we have $(xy, x^4 + y^4) = 1$ and hence $xy$ divides $d$. As $x^4 + y^4 \ge 17$ and $p$ is a prime, we get $d = xy$ and $p = x^4 + y^4$. Since one of $x$ and $y$ is even and the other one is odd, $p = x^4 + y^4 \equiv 1 \pmod 8$.

If $m = 2dxy$, then $p = d(x^2 - y^2) + \frac{8dx^2y^2}{x^2 - y^2}$. Since $\frac{8dx^2y^2}{x^2 - y^2}$ is an integer and $(xy, x^2 - y^2) = 1$, we have $x^2 - y^2$ divides $8d$. On the other hand, as $p$ is a prime number, $d$ and $\frac{8d}{x^2 - y^2}$ are relatively prime. Thus, $x^2 - y^2$ is a multiple of $d$. Therefore, $x^2 - y^2 = d, 2d, 4d$ or $8d$.
If $x^2 - y^2 = d$, then as $p = d^2 + 8x^2y^2$ is a prime, $d$ is odd and $p \equiv 1 \pmod 8$.
If $x^2 - y^2 = 2d$, then $p = 2(d^2 + 2x^2y^2)$ is not a prime number.
If $x^2 - y^2 = 4d$, then $p = 2(2d^2 + x^2y^2)$ is not a prime number.

If $x^2 - y^2 = 8d$, then $p = 8d^2 + x^2y^2$ is a prime number implies that both $x$ and $y$ are odd numbers and consequently $p \equiv 1 \pmod 8$.