This problem explores the properties of quadratic functions.

(1) When $a = -1$, we substitute $a$ into the function to get $f(x) = x^2 - 2x + 2$. Completing the square, we have $f(x) = (x - 1)^2 + 1$. Since we are given $x \in [-5,5]$, this is a shifted parabola that opens upwards with the vertex at $(1,1)$.

Thus, the minimum value of $f(x)$ occurs at the vertex of the parabola:
$$f(x)_{\text{min}} = f(1) = (1 - 1)^2 + 1 = \boxed{1}$$

The maximum value of $f(x)$ will occur at one of the endpoints of the interval, since the vertex is inside the interval and the function is symmetric about the line $x = 1$. We evaluate the function at $x = -5$:
$$f(x)_{\text{max}} = f(-5) = (-5 - 1)^2 + 1 = 36 + 1 = \boxed{37}$$

(2) For $f(x)$ to be monotonic on $[-5,5]$, the vertex of the parabola $(x+a)^2 - a^2 + 2$ must not be inside the interval $[-5,5]$. The vertex of the parabola is at $x=-a$. For the function to be monotonic, $-a$ must satisfy either $-a \le -5$ or $-a \ge 5$. Converting these inequalities, we find the range of $a$:
$$a \ge 5 \quad \text{or} \quad a \le -5$$

Therefore, the values of $a$ for which $f(x)$ is monotonic on $[-5,5]$ are:
$$\boxed{a \in (-\infty, -5] \cup [5, \infty)}$$