a. It is sufficient to prove that $1+1=0$. Consider $x$ noninvertible, and let $m \in \mathbb{N}^*$, such that $x = x^2 + x^3 + \dots + x^{2^m}$, and $y = x + x^2 + \dots + x^{2^m-1}$. Evidently, $xy = x$, so $xy^k = x$, for any $k \in \mathbb{N}^*$. Because $x$ is not invertible it follows that $y$ is also, so there is $p \in \mathbb{N}^*$, such that $-y = (-y)^2 + (-y)^3 + \dots + (-y)^{2^p-1} + (-y)^{2^p} = y^2 - y^3 + \dots - y^{2^p-1} + y^{2^p}$. As a consequence $-x = -xy = xy^2 - xy^3 + \dots - xy^{2^p-1} + xy^{2^p} = x - x + \dots - x + x = x$, i.e., $x+x=0$.

Consider $x$ non zero and non-invertible. As $2x = 0$, it follows that $2$ is non-invertible, so $2+2=0$ and $2 = 2^2 + 2^3 + \dots + 2^{2^m}$, where $m \in \mathbb{N}^*$. The equality $2+2=0$ implies $2^2 = 0$, giving $2^k = 0$, for any $k \ge 2$. So, $2 = 2^2+2^3+\dots+2^{2^m} = 0$.

b. Let $x$ be a non-invertible element of $A$ and $m \in \mathbb{N}^*$, such that $x = x^2 + x^3 + \dots + x^{2^m}$. Then $x^2 = x^3 + x^4 + \dots + x^{2^m+1}$. As $1+1=0$, by summing up the two equalities we get $x^{2^m+1} = x$.

Equalities $1+1=0$ and $x^{2^m+1} = x$ imply $(x^2+x)^{2^m} = x^{2^m+1} + x^{2^m} = x^{2^m+1} \cdot x^{2^m-1} + x^{2^m} = x \cdot x^{2^m-1} + x^{2^m} = x^{2^m} + x^{2^m} = 0$.

We will show that $x^2+x=0$, which concludes the proof. Let $y = x^2+x$ and take $k$ the smallest positive integer such that $y^k = 0$ – the existence of $k$ is assured by $y^{2^m} = 0$. In case $k > 1$, the element $y^{k-1}$ is non-invertible in $A$ and there is $n \in \mathbb{N}^*$ such that $y^{k-1} = (y^{k-1})^{2^n+1} = y^{(k-1)(2^n+1)}$. Because $(k-1)(2^n+1) \ge k$, we infer $y^{(k-1)(2^n+1)} = 0$, so $y^{k-1} = 0$ – in contradiction with the minimality of $k$. We conclude $k=1$ and $y=0$.