Solution:

a) Since $\lim_{x \rightarrow +\infty} f'(x) = +\infty$ and $\lim_{x \rightarrow -\infty} f'(x) = -\infty$, it is enough to show that the local minimum $m$ of $f'(x)$ is positive. Since the equation $f''(x) = 0$ has two real roots $x_{1} > x_{2}$, it follows that $m = f'(x_{1}) > 0$. Now it is easy to check that $x_{1} \in (-1 ; 0)$ and $m > 0$.

b) It follows from a) that the equation $f'(x) = 0$ has a unique real root. Since $\lim_{x \rightarrow +\infty} f(x) = \lim_{x \rightarrow -\infty} f(x) = +\infty$ and $f(0) < 0$, we conclude that the equation $f(x) = 0$ has exactly two real roots. To find them, set $y = 2003$ and consider $f(x) = 0$ as a quadratic equation with respect to $y$. We have
$$y_{1,2} = \frac{x \pm x(4x + 3)}{2}$$
and then either $2y = x - x(4x + 3)$ or $2y = x + x(4x + 3)$. For $y = 2003$ the first equation has no real roots and the second one has two real roots $x_{1,2} = \frac{-1 \pm \sqrt{4007}}{2}$.