a) The first polynomial, i.e. $2x^2-1$, can be obtained by applying the following sequence of steps:
$$x^2 - 2x - 1 \rightarrow -x^2 - 2x + 1 \xrightarrow{d=-1} -x^2 + 2 \rightarrow 2x^2 - 1.$$

b) Notice that the prescribed steps leave the discriminant unchanged: the discriminant of $cx^2+bx+a$ is $b^2-4ca = b^2-4ac$, whereas the discriminant of
$$a(x+d)^2 + b(x+d) + c = ax^2 + (2ad+b)x + (ad^2+bd+c)$$
equals
$$(2ad+b)^2 - 4a(ad^2+bd+c) = 4a^2d^2 + 4abd + b^2 - 4a^2d^2 - 4abd - 4ac = b^2 - 4ac.$$
The discriminant of the initial polynomial $x^2-2x-1$ is $8$, while the discriminant of $2x^2-x-1$ equals $9$. Thus, Branko cannot obtain the polynomial $2x^2-x-1$ in any number of allowed steps.

Remark: Consider the prescribed steps as operations T1 and T2 on triples of real numbers:
$$(a, b, c) \xrightarrow{T1} (c, b, a) \quad \text{and} \quad (a, b, c) \xrightarrow{T2} (a, 2ad+b, ad^2+bd+c).$$
Given any two quadratic polynomials $f(x)$ and $g(x)$ with equal discriminants, we can obtain a sequence of steps transforming $f(x)$ into $g(x)$: first, we apply T2 to adjust the constant term of $f(x)$, making it equal to the leading coefficient of $g(x)$. After that, we can use T1 to swap the leading coefficient with the constant term. Finally,
using T2, we can adjust the new constant term making it equal to the constant term of $g(x)$.
This last step guarantees that the resulting polynomial will have the same constant term as $g(x)$. The leading coefficient is also equal to the one appearing in $g(x)$, because it remains unchanged by T2. Finally, the linear term also matches the one appearing in $g(x)$: it is uniquely determined (up to sign) by the discriminant and the remaining two coefficients. If necessary, we can change its sign by applying T2 with $d = -b/a$.