Solution:

a) The equation (1) can be written as $9^{x} = 2^{x}$, i.e. $\left(\frac{9}{2}\right)^{x} = 1$ and $x = 0$.

b) We plug the only solution $x = 0$ of (1) in (2) and obtain $|a-1| = 1-a \Longleftrightarrow a \leq 1$. In this case $0$ is a solution of (2) and we have to decide when (2) has no other solution(s).

Set $5^{x} = t$, $t > 0$. Then we have to find all $a \leq 1$ such that the equation
$$a t^{2} + (1-a) t - 1 = 0$$
has not positive roots different from $1$. For $a = 0$ the only root of (3) is $t = 1$ and the condition is satisfied. For $a \neq 0$ the roots of (3) are $1$ and $-\frac{1}{a}$. Therefore the condition is satisfied if and only if $-\frac{1}{a} = 1$ or $-\frac{1}{a} < 0$. Hence $a = -1$ or $a > 0$ and we conclude that $a \in [0, 1] \cup \{-1\}$.