a.
The sequence starts as follows.
$$\begin{aligned} a_1 &= \frac{10}{11}, & a_2 &= \frac{12}{14} = \frac{6}{7}, & a_3 &= \frac{8}{10} = \frac{4}{5}, & a_4 &= \frac{6}{8} = \frac{3}{4}, \\ a_5 &= \frac{5}{7}, & a_6 &= \frac{7}{10}, & a_7 &= \frac{9}{13} \end{aligned}$$
It seems that the last simplification occurred at $a_4$. With induction to $n$, we will prove that there is no simplification for all $n \ge 5$. At the same time, we will prove that $a_n = \frac{1+2(n-3)}{1+3(n-3)}$ for all $n \ge 5$.
For $n = 5$, the statement is true, because $a_5 = \frac{5}{7} = \frac{1+2(5-3)}{1+3(5-3)}$ and this fraction $\frac{5}{7}$ cannot be simplified further. Now suppose the statement is true for $n = k - 1$. Consider $n = k$. Because there has been no simplification for $a_{k-1}$, the numerator of $a_{k-1}$ equals $1 + 2(k - 4)$ and the denominator equals $1 + 3(k - 4)$. Then the number $a_k$ is defined as $\frac{1+2(k-4)+2}{1+3(k-4)+3} = \frac{1+2(k-3)}{1+3(k-3)}$.
We will argue by contradiction that there is no simplification here. Namely, suppose there is an integer $d > 1$ such that both $1 + 2(k - 3)$ and $1 + 3(k - 3)$ are divisible by $d$. In particular, $3 \cdot (1 + 2(k - 3)) - 2 \cdot (1 + 3(k - 3)) = 1$ will also be divisible by $d$. This gives a contradiction, and the proof by induction is finished. $\square$

b.
We will show that there must be a simplification at some point. Indeed, suppose there is no simplification. Just like in part (a), we can show by induction that $a_n = \frac{97+2n}{97+3n}$. In particular, we see that $a_{97}$ is not a simplified fraction, because both the numerator and denominator are divisible by 97, and that is a contradiction. $\square$

c.
You can use $c = 7$ or $c = 27$, for example. Then we get the sequences
$$\frac{7}{8}, \frac{9}{11}, \frac{11}{14}, \frac{13}{17}, \frac{15}{20} = \frac{3}{4}$$
and
$$\frac{27}{28}, \frac{29}{31}, \frac{31}{34}, \frac{33}{37}, \frac{35}{40} = \frac{7}{8}.$$
$\square$