a) We seek for seven consecutive integers $a, a+1, a+2, a+3, a+4, a+5, a+6$ such that $7^2 = a + a + 1 + a + 2 + a + 3 + a + 4 + a + 5 + a + 6$. The equality gives $49 = 7a + 21$, hence $a = 4$. As $7^2 = 4 + 5 + 6 + 7 + 8 + 9 + 10$, number $7$ is lucky.

b) Suppose there exist ten consecutive integers, namely $a, a+1, a+2, \dots, a+9$, such that $10^2 = a + a + 1 + a + 2 + \dots + a + 9$. The equality gives $100 = 10a + 45$, with no integer solution. Number $10$ is not lucky.

c) We claim that a number $m$ is lucky if and only if $m$ is odd. Suppose $m$ is a lucky number and write $m^2 = (a+1) + (a+2) + (a+3) + \dots + (a+m)$ for a given positive integer $a$. Then
$$m^2 = m \cdot a + (1 + 2 + 3 + \dots + m) \Leftrightarrow m^2 = m \cdot a + \frac{m \cdot (m+1)}{2},$$
hence $m = 2a + 1$, so $m$ is odd. Conversely, let $m = 2k + 1$ be an odd number. As $(2k+1)^2 = (k+1) + (k+2) + \dots + (3k+1)$, the number $m$ is lucky.

We conclude noticing that the product of two odd numbers is odd.