Asymptotics and logarithmic exponents
First, notice that $3 &gt; \ln 7$. You can prove this formally using the Taylor expansion of $e^x$:
$$
e^3 = \sum_{n=0}^\infty \frac{3^n}{n!} \geq 1 + 3 + \frac{9}{2} = 8.5 &gt; 7.
$$
Monotonicity of $\ln$ implies that indeed $3 &gt; \ln 7$.

Armed with this knowledge, we deduce that for $n \geq 1$,
$$
n^3 = n^{3 - \ln 7} n^{\ln 7} \geq n^{\ln 7}.
$$
This shows that $n^{\ln 7} = O(n^3)$, using the definition of big O.

For your other question, try using $8^{\log_2 n} = n^3$ and induction. The idea is that as you increase $n$ by 1, the value of $n^3$ hardly increases, whereas $5^n$ increases by a factor of 5.

Finally, let me say that there is absolutely no reason to be so formal. It's enough to go through the formalities once, just to get convinced that they work, and afterwards you can just use well-known facts such as $n^k = o(c^n)$ for all $c &gt; 1$.