Solution:

a) It follows by a direct verification.

b) Writing the equation in the form
$$(x + a)\left((a^{2} + 4a + 2)x^{2} + (1 - a)x + a\right) = 0$$
we get that $a < 0$. Moreover, the quadratic polynomial in (3) must have two distinct real zeros, i.e.
$$D = (1 - a)^{2} - 4a(a^{2} + 4a + 2) > 0 \Longleftrightarrow (a + 1)(-4a^{2} - 11a + 1) > 0.$$
Solving this inequality and having in mind that $a < 0$, we get that
$$a \in \left(-\infty, \frac{-11 - \sqrt{137}}{8}\right) \cup (-1, 0)$$
The roots of the quadratic polynomial in (1) are positive if and only if
$$\frac{a}{a^{2} + 4a + 2} > 0 \quad \text{and} \quad -\frac{1 - a}{a^{2} + 4a + 2} > 0.$$
Then $a \in (-2 - \sqrt{2}, -2 + \sqrt{2})$ and using (4), we obtain that
$$a \in \left(-2 - \sqrt{2}, \frac{-11 - \sqrt{137}}{8}\right) \cup (-1, -2 + \sqrt{2})\text{.}$$
It remains to see when $-a$ is a zero of the quadratic polynomial in (3). We have that
$$(a^{2} + 4a + 2)(-a)^{2} + (1 - a)(-a) + a = 0 \Longleftrightarrow a^{2}(a^{2} + 4a + 3) = 0$$
and hence $a = -3, -1, 0$. So, the answer of the problem is
$$a \in (-2 - \sqrt{2}, -3) \cup \left(-3, \frac{-11 - \sqrt{137}}{8}\right) \cup (-1, -2 + \sqrt{2})$$