I am solving following problem in the mentioned title.
Q.1 I didn't get direction from last two statements in Hint.
What I did? I think the hypothesis n is prime power and cyclic implies that there exists irreducible character $\chi$ whose kernel intersects $H$ trivially; beyond this, this hypothesis has no use further. Then I have to look at use of second hypothesis: $m$ is maximum order of elements of $G/H$. I didn't get direction how to use it? And finally, Problem 9.2(b) says that if $A$ is an abelian subgroup of $G$, then any character has degree $\leq|G:A|$. In problem, the required abelian subgroup is perhaps $H$; but we don't know $|G:H|$ from hypothesis.
Q.2 For what types of groups this problems tells something about group? I mean, any simple application of this problem?
Since $H$ is non-trivial cyclic, one can find a linear $\mu \in Irr(H)$, with $ker(\mu)=1$. In fact, these linear characters are precisely those which generate $Irr(H)$ as a cyclic group. Now take an irreducible constituent $\chi \in Irr(G)$, with $[\mu^G,\chi] \neq 0$. Since $H \subseteq Z(G)$, we must have $\chi_H=\chi(1)\mu$. Observe that $ker(\chi_H)=H \cap ker(\chi)=ker(\mu)=1$.
Putting $\lambda = det(\chi)$ (see Problem(2.3) of Isaacs' book as he indicates), one creates a linear character of $G$. Since $H \subseteq G'$, we must have $\lambda(h)=1$ for all $h \in H$. In particular, we can calculate this value on a generator of $H$, say $H=\langle x \rangle$, then $1=\lambda(x)= det(\chi(x))=det(\chi(1)\mu(x))=\mu(x)^{\chi(1)}$. Since $\mu(x)$ is a primitive $n$-root of unity, this implies that $n$ divides $\chi(1)$.
Further, if $m$ is the maximum of the orders of the elements of $G/H$, pick $\bar{g} \in G/H$ with $o(\bar{g})=m$. Put $A=H\langle g \rangle$. Then $A$ is an abelian subgroup of $G$, since $H$ is central. By (2.9)(b) of Isaacs' book, we have $\chi(1) \leq [G:A]$ (for a proof see here). Hence, putting it all together, $|G| \geq n|A|=n \cdot \frac{|H| \cdot |\langle g \rangle|}{|H \cap \langle g \rangle|}=n \cdot \frac{|H|\cdot|\langle g \rangle|}{|\langle g^m \rangle|} \geq n^{2}m$. Note that $H \cap \langle g \rangle = \langle g^m \rangle$, because $m$ is chosen to be maximal among the orders of elements $G/H$.
There is for example a class of extra-special $p$-groups ($p$ prime) of order $p^3$, with $G'=Z(G)$ cyclic of order $p$, and $G/Z(G)$ isomorphic to $C_p \times C_p$, satisfying this exercise.