You wish to prove that $P(\limsup_n \frac{Y_n}{\lambda \log n} = 1) =1$. It suffices to prove that $$P(\limsup_n \frac{Y_n}{\lambda \log n} > 1) = P(\limsup_n \frac{Y_n}{\lambda \log n} < 1)=0$$
Let $\displaystyle X_n = \frac{Y_n}{\lambda \log n}$.
Note that $\displaystyle \limsup_n \frac{Y_n}{\lambda \log n} > 1 = \{w\in \Omega, \limsup_n \frac{Y_n(w)}{\lambda \log n} > 1\} = \bigcup_N \bigcap_n \bigcup_{k\geq n}\left(X_k> 1+\frac 1N \right)$.
It suffices to prove that $\forall N, P\left(\bigcap_n \bigcup_{k\geq n}\left(X_k> 1+\frac 1N \right) \right)=0$.
But $P\left(\bigcap_n \bigcup_{k\geq n}\left(X_k> 1+\frac 1N \right) \right)=P\left(\limsup_n \left( X_n> 1+\frac 1N\right) \right)$.
By Borel-Cantelli, it suffices to prove that $\sum_n P\left( X_n> 1+\frac 1N\right)$ converges. Note that $$P\left( X_n> 1+\frac 1N\right) = P\left(Y_n > \lambda \left( 1 + \frac 1N \right)\log n\right)=\frac{1}{n^{1+\frac 1N}}$$ and convergence follows.
This proves $P(\limsup_n \frac{Y_n}{\lambda \log n} > 1) =0$.
Note that $$ \begin{aligned}[t] \left(\limsup_n \frac{Y_n}{\lambda \log n} \right)< 1 &= \{w\in \Omega, \limsup_n \frac{Y_n(w)}{\lambda \log n} < 1\} \\&\subset \bigcup_n \bigcap_{k\geq n}\left(\frac{Y_k}{\lambda \log k} < 1\right)\\ &= \liminf_n \left( \frac{Y_n}{\lambda \log n} < 1\right) \end{aligned}$$
Since $\liminf_n \left( \frac{Y_n}{\lambda \log n}<1 \right)=\left(\limsup_n \left(\frac{Y_n}{\lambda \log n}>1\right) \right)^c $, it suffices to prove that $$P\left( \limsup_n \left( \frac{Y_n}{\lambda \log n}>1\right)\right)= 1$$
As expected we make use of the second Borel Cantelli lemma (after an independence assumption is added on the $Y_k$) : $$P\left( \frac{Y_n}{\lambda \log n}>1\right) =P\left( Y_n>\lambda \log n\right)= \frac 1n $$ and the sum diverges, proving the claim.
For the second part of your question, I suppose you can proceed similarly.
