a) The equation is written equivalently $[x]^2 - [x] = \{x\} - 0.99$, thus $\{x\} - 0.99 \in \mathbb{Z}$.
Since $0 \le \{x\} < 1$, we deduce that $-0.99 \le \{x\} - 0.99 < 0.01$ therefore $\{x\} - 0.99 = 0$ so $\{x\} = 0.99$. Also, from $[x]^2 - [x] = 0$ we deduce that $[x] = 0$ or $[x] = 1$, so $x \in \{0.99; 1.99\}$.

b) The equation is written equivalently $[x]^2 - [x] = \{x\} + a$. We assume, by absurdity, that there is $a \le -1$ for which the equation has real solutions. Then, since $\{x\} < 1$, $[x]^2 - [x] = \{x\} + a < 0$.
Denoting $[x] = y \in \mathbb{Z}$, we have $y(y - 1) < 0$, so $y \in (0, 1)$, absurd.