$k(n)=[\frac{\ln a_{n}}{\ln A }]+1$. It prove all.

First, there are a few typos to be fixed. $1)$ The sequence must be unbounded from above. $2)$ Part $(b)$ preamble should be $A^{f(n)}$, rather than $A^{k(n)}$. We will proceed as follows. $a)$ We will first explicitly recover $k(n)$, and prove that it is surjective and non-decreasing. From the preamble, $$ \log a_n < k(n)\log A \leq \log a_n + \log A \implies \frac{\log a_n}{\log A}<f(n)\leq \frac{\log a_n}{\log A}+1. $$Hence, $$ \boxed{f(n)=\left\lfloor\frac{\log a_n}{\log A}+1\right\rfloor}, $$since $f(n)$ is integer-valued. Clearly, $f(n)$ is uniquely defined, and non-decreasing. For surjectivity, note that $f(n)=k$ if and only if, $A^{k-1}\leq a_n < A^k$. Now, we will show that for every given $k$, there exists an $n$ such that this will hold (and this is where unboundedness comes into the picture). Fix a $k$ and let $n_k$ be the smallest integer such that $a_{n_k}\geq A^{k-1}$. Since $a_{n_k-1}<A^{k-1}$, we have $a_{n_k}\leq Aa_{n_k-1}<A^k$, and therefore, $f(a_{n_k})=k$. $b)$ For this part, since $f(\cdot)$ can take every value at most $m$ times, we have $f(m+1)\geq 2$, $f(2m+1)\geq 3$, etc. Hence, $f(nm+1)\geq n+1$, for every $n\in\mathbb{N}$. Now, $$ f(nm+1)\geq n+1 \implies \frac{\log a_{nm+1}}{\log A}\geq n \implies a_{nm+1}\geq A^n=\left(A^{\frac{n}{nm+1}}\right)^{nm+1}. $$Hence, it suffices to take $C=A^{t}$, where, $t$ is defined as, $$ t=\inf\left\{\frac{n}{nm+k}:n,k\in\mathbb{N},k\leq m\right\}, $$which is (easy to check) equal to $1/2m$.