(a) We take $a = 4k - 3$, $k \ge 2$, and then take $m = 2k - 1$, $n = k - 1$, we obtain
$$(m^2+n)(n^2+m) = ((2k-1)^2+(k-1))((k-1)^2+(2k-1)) = (4k-3)k^3 = a(m-n)^3.$$
Therefore $5, 9, \dots, 2009, 2013$ are friendly and $\{1, 2, \dots, 2013\}$ contains at least 503 friendly integers.

(b) We prove that $a = 2$ is not friendly. We consider the equation $(m^2+n)(n^2+m) = 2(m-n^3)$ when $a = 2$, and write the left side in the form of a difference of squares:
$$\frac{1}{4}((m^2+n+n^2+m)^2 - (m^2+n-n^2+m)^2) = 2(m-n)^3. \quad (1)$$
Since $m^2+n-n^2-m = (m-n)(m+n-1)$, we can obtain
$$(m^2+n+n^2+m)^2 = (m-n)^2(8(m-n) + (m+n-1)^2).$$
Therefore $8(m-n) + (m+n-1)^2$ is a perfect square. Clearly $m > n$, so there exists an integer $s \ge 1$ such that
$$(m+n-1+2s)^2 = 8(m-n) + (m+n-1)^2.$$
Expanding the above we get $s(m+n-1+s) = 2(m-n)$. Since $m+n-1+s > m-n$, we have $s < 2$. So the only possibility is $s = 1$ and $m = 3n$. Substituting $m = 3n$ into (1) we find $27n^3 = 16n^3$, giving $n = m = 0$, a contradiction. Therefore $a = 2$ is not friendly.