Asymptotics claims (little o and exponentiation) I have been stuck on this for hours and would really appreciate some help with it. I have to check if those statements imply the given outcome.

a) $f(n)=o(g(n))$ implies that $3^{f(n)}=o(3^{g(n)})$
b) $3^{f(n)}=o(3^{g(n)})$ implies that $f(n) = o(g(n))$

I am really having problem with this and would appreciate your help. here's what I did:
a) seems to be a true claim.since if $f(n)=o(g(n))$ then $0\leq f(n)<cg(n)$. so in this case $3^{f(n)} <c\cdot 3^{g(n)}$, which makes it a true claim.
b) here i really don't know. if $3^{f(n)}=o(3^{g(n)})$ then it should mean $3^{f(n)}< 3^{Cg(n)}=(3^C)^{g(n)}$. I don't know how to deduct from here whether it is a true claim or not.
Thank you very much for your help.

So, the main takeaway here is be careful with exponentiation and asymptotics. It is not quite always what you'd expect.

a). This is false... first, a counterexample: $f(n) = \frac{1}{n}$, $g(n)=1$. We clearly have $f(n) = o(g(n))$ when $n\to\infty$. Yet,
$$
\frac{3^{f(n)}}{3^{g(n)}} = 3^{\frac{1}{n}-1} \xrightarrow[n\to\infty]{} \frac{1}{3} > 0
$$
so we actually have $3^{f(n)} = \Theta(3^{g(n)})$. 
Why? $f(n)=o(g(n))$ does not mean much for $f(n)-g(n)$ (it could go to $-\infty$, or as above... to $0$. But that's the exponent of the ratio $\frac{3^{f(n)}}{3^{g(n)}}$.
b). This is false again. $3^{f(n)} = o(3^{g(n)})$ (when $n\to\infty$) is equivalent to $3^{f(n)-g(n)} = \frac{3^{f(n)}}{3^{g(n)}} = o(1)$, that is 
$$
\lim_{n\to\infty} 3^{f(n)-g(n)}  = 0\,.
$$
By continuity of the logarithm, taking $\log$'s we get 
$$
\lim_{n\to\infty} (f(n)-g(n))  = -\infty\,.
$$
This does not imply $f(n) = o(g(n))$ either, only that the difference goes to $\infty$ (instead of the ratio going to $1$). For a counterexample:
$$
f(n) = n^2, \qquad g(n) = n^2+n\,.
$$
