Let $\mathbb F_{27}$ be the finite field of order $27$. Let $A_{a}=\{1,1+a,1+a+a^2,\dots\}$. Then which of the following are true?
The number of $a\in\mathbb F_{27}$ such that $\operatorname{card}(A_{a})=26$ equals to $12$.
$0 \in A_{a} \iff a$ is nonzero.
$\operatorname{card}(A_1)=27$
$\bigcap A_{a}$ is a singleton.
For 2 one way is very obvious. If $0=a$ then $0=1+a+a^2+... +a^k$ for some $k <= 27$.Put a=0 and we get $1=0$, which is a contradiction. For the other way round I am unable to proceed.
For 4 I have no ideas how to proceed.
If $a=0$, then $A_0=\{1\}$ and if $a=1$, then $A_{1}=\{0,1,2\}$. For the following let us assume that $a \neq 0,1$.
An element of the form $1+a+a^2+\dotsb+a^{k-1}=(a^k-1)(a-1)^{-1}$ (note: $a-1$ is invertible in $\Bbb{F}_{27}$, because $a \neq 1$). Let us call the element $(a-1)^{-1}=b$. Since $a \in \Bbb{F}_{27}-\{0,1\}$, so $a^{26}=1$. Observe that we can write $$A_a=\{1,b(a^2-1),b(a^3-1), \ldots, b(a^{25}-1),0\}.$$ For $|A_a|=26$, we want all the elements in this list to be distinct.
Ques: When will $b(a^k-1)=b(a^j-1)$?
For this to occur, $a^{k-j} \equiv 1$. This means the order of $a$ is less than or equal to $k-j$ and $k-j<26$. Thus if order of $a$ is exactly $26$ then we will not have these collisions in the set $A_a$. Same as saying that $a$ must be a primitive root (or generator) of $\Bbb{F}_{27}^{\times}$.
So the number of such $a'$s is $\color{red}{\phi(26)=\phi(2)\phi(13)=12.}$
Part(2): If $a \neq 0$, then the order of $a$ is some positive integer $k$. In which case the element $b(a^k-1)=0$, thus $0 \in A_{a}$ for $a \neq 0$. I have already shown above that $0 \not\in A_0$.
Part(3): is false as shown above.
Part(4): is trivially true because $A_0=\{1\}$ has only one element.