Let $\,\,\,\sigma$, $\tau$ $\in S_8\,\,\,\,$ s.t. $\,\,\,\sigma = (123)(456)\,\,\,$ and $\,\,\,\tau=(14)(25)(36)(78)$
Let $\,\,\,\phi$, $\delta$, $\lambda \in S_6\,\,\,$ s.t. $\,\,\,\phi = (12)(34)(56)$, $\,\,\,\delta = (34)(56)\,\,\,$ and $\,\,\,\lambda = (126)$
Say if $\,\,\,\langle \sigma, \tau\rangle \,\,\,$ is cyclic in $S_8$ and if $\,\,\,\langle\phi \delta, \lambda\rangle \,\,\,$ is cyclic in $S_6$ and prove that.
About $\,\,\langle \sigma,\tau\rangle \,$, it is easy to observe that $\,\,\langle \sigma\rangle\langle \tau\rangle \,\, = \,\,\,\langle\sigma, \tau\rangle \,$, furthermore $\,\,\,\langle \tau\rangle \leqslant N_{S_8}(\langle\sigma\rangle)\,\,\,$ and viceversa, eventually $\,\,\,(|\sigma|, |\tau|)=1\, \,\,\,\Rightarrow \,\,\,\,\langle\sigma\rangle \cap\langle \tau\rangle \,\,= \{1\}\,\,\,$; therefore I can say: $\,\,\,\langle\sigma, \tau\rangle \,\,\, = \,\,\,\langle\sigma\rangle\langle\tau\rangle \,\,\, \simeq \,\,\,\langle\sigma\rangle \times\langle\tau\rangle \,\,\, \simeq \,\,\,C_3\times C_2\,\,\, \simeq\,\,\, C_6$
Regarding $\,\,\,\langle\phi \delta, \lambda\rangle \,\,\,$ I would say also yes, but I could not find a proper proof, if I am right, any suggestion?
Note that $\phi\delta=(1\ 2)$, so $\phi\delta$ and $\lambda$ only permute the three elements $1$, $2$ and $6$. In this way we see that $\langle\phi\delta,\lambda\rangle$ is isomorphic to a subgroup of $S_3$. Can you now determine whether this group is cyclic or not?
As for the subgroup $\langle\sigma,\tau\rangle$, a more explicit argument would be to show that $$\sigma^3=\operatorname{id},\qquad\tau^2=\operatorname{id},\qquad\sigma\tau=\tau\sigma.$$ It follows that $(\sigma\tau)^4=\sigma$ and $(\sigma\tau)^3=\tau$, so $\langle\sigma,\tau\rangle=\langle\sigma\tau\rangle$.