The roots of the polynomial $p(x) = x^2 + a x + b$ are $x_1 = \frac{-a + \sqrt{a^2 - 4b}}{2}$ and $x_2 = \frac{-a - \sqrt{a^2 - 4b}}{2}$. They have to be different, so $a^2 - 4b \neq 0$. From here we get
$$x_1^2 - \frac{1}{2} = \frac{(a^2 - 2b - 1) - a \sqrt{a^2 - 4b}}{2}$$
and
$$x_2^2 - \frac{1}{2} = \frac{(a^2 - 2b - 1) + a \sqrt{a^2 - 4b}}{2}.$$
Since these are the roots of the polynomial $q(x)$, they have to satisfy the equation $x^2 + (a^2 - \frac{1}{2}) x + b^2 - \frac{1}{2} = 0$. Inserting both into this equation and rearranging we get
$$(4a^4 - 12a^2b + 8b^2 - 5a^2 + 6b) - (4a^3 - 4ab - 3a)\sqrt{a^2 - 4b} = 0$$
and
$$(4a^4 - 12a^2b + 8b^2 - 5a^2 + 6b) + (4a^3 - 4ab - 3a)\sqrt{a^2 - 4b} = 0.$$
So,
$$4a^4 - 12a^2b + 8b^2 - 5a^2 + 6b = 0$$
and
$$(4a^3 - 4ab - 3a)\sqrt{a^2 - 4b} = 0.$$
Since $a^2-4b \neq 0$ the second equation implies that $(4a^3-4ab-3a) = a(4a^2-4b-3) = 0$. If $a=0$, then the first equation implies that $8b^2+6b = 2b(4b+3) = 0$. Since $a^2-4b \neq 0$ we have $b \neq 0$, so $b = -\frac{3}{4}$. If $a \neq 0$, then $4a^2 - 4b - 3 = 0$, or $b = a^2 - \frac{3}{4}$. Using this in the first equation we get $-2a^2 = 0$, or $a = 0$, which is a contradiction. So, $a = 0$ and $b = -\frac{3}{4}$.