Give an example of each or state that the request is impossible. For a

Step 1
Take sequence ${\displaystyle x_n={(-1)}^n}$ where 1 occurs indefinitely but does not converge to 1.
This is because if we choose ${\displaystyle ϵ\le 2}$, given any ${\displaystyle N\in \mathbb{N}}$, where exists ${\displaystyle n=2N+1}$ such that
${\displaystyle |{\left(1\right)}^n-1|=2\ge ϵ}$
Step 2
b) Suppose ${\displaystyle x_n}$ is a convergent sequence with infinitely many 1 but converges to ${\displaystyle x\ne 1}$. Let ${\displaystyle ϵ=|1-x|,}$ then for any ${\displaystyle N\in \mathbb{N}}$, there exist ${\displaystyle n\ge N}$ such that ${\displaystyle x_n=1}$. So
${\displaystyle |x_n-x|=|1-x|\ge ϵ}$
So this is not possible
Step 3
c) Consider a sequence ${\displaystyle \left(x_n\right)}$ given by
$x_n=\{\begin{array}{ll}r& ifn=\frac{r(r+1)}{2}-1\text{for some}r\in \mathbb{N}\\ 1& \text{otherwise}\end{array}$
(or put any thing in place of r other than 1)
The sequence contains n consecutive 1's. Some of the terms in order are
${\displaystyle {\left(x_n\right)}_{n\in \mathbb{N}}=(1,2,1,1,5,1,1,1,9,1,1,1,1,14,\cdots )}$
This is constructed this way
$\begin{array}{cc}1& 2\\ 1& 1& 5\\ 1& 1& 1& 9\\ \cdots \end{array}$

Step 1
a) Consider the sequence ${\displaystyle x_n={(-1)}^n}$
Here, 1 appears infinitely many times and does not converge to 1.
Thus, on choosing ${\displaystyle ϵ\le 2}$, given any ${\displaystyle N\in \mathbb{N}}$, there exist ${\displaystyle n=2N+1}$ such that ${\displaystyle |{\left(1\right)}^n-1|=2\ge ϵ}$
Step 2
b) Consider a convergent sequence ${\displaystyle \left(x_n\right)}$ with infinitely many 1 but it converges to ${\displaystyle x\ne 1}$.
Let ${\displaystyle ϵ=|1-x|}$ then for any ${\displaystyle N\in \mathbb{N}}$, there exist ${\displaystyle n\ge N}$ such that ${\displaystyle x_n=1}$. So,
${\displaystyle |x_n-1|=|1-x|\ge ϵ}$
Hence, this is not possible.
Step 3
c) Consider a sequence ${\displaystyle \left(x_n\right)}$ given by
$$x_n=\{\begin{array}{ll}r& ifn=\frac{r(r+1)}{2}-1\text{for some}r\in \mathbb{N}\\ 1& \text{otherwise}\end{array}$$
There exist n consecutive 1's occur in the sequence. Some of the terms in the sequence will be
$$(x_n{)}_{n\in \mathbb{N}}=(1,2,1,1,5,1,1,1,9,1,1,1,1,14,\cdots )$$
Which gives a contradiction.
That is,
$\begin{array}{cc}1& 2\\ 1& 1& 5\\ 1& 1& 1& 9\\ \cdots \end{array}$
Answer: it's impossible.

Step 1
a) Consider the sequence
$\{a_n\}=\{(-1{)}^n\}=\{-1,1,-1,1,\cdots \}$
$a_n$ has an infinite number of ones. Indeed, if n is even, $(-1{)}^n=1$, and there are infinite even numbers, hence infinite ones. However, $\{a_n\}$ does not converge to one, because $\{a_n\}$ is divergent (it oscillates between 1 and -1).
b) This is impossible. We will use the definition of limit to see why.
Let $ϵ>0$. Suppose that $\{a_n\}$ has an infinite number of ones, and suppose that $\{a_n\}$ converges to L. Then, there exists some M>0 such that if $m\ge M$ then $|a_n-L|<ϵ$. Now, there exist infinite natural numbers $n\ge M$ such that $a_n=1$ by hypotheses. Then $|1-L|<ϵ$
Hence L and 1 get arbitrarily close, therefore L=1
c) It is possible to construct such a sequence:
$\{a_n\}=\{1,0,1,1,0,1,1,1,0,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,1,0,\cdots \}$
even if there is not a simple explicit formula. The sequence is divergent because it has infinite zeroes and infinite ones (thus L should be either 0 or 1, but it can't be both).