Q1: Here is where my first problem arrived; I was not given a relation for T(1) so I can not finish with substitution, right? Perhaps my professor made a typo and meant to put T(1) = 1?
No, your professor was correct in providing $T(3) = 1$, given $n = 3^{k}$. As you may have noticed when using substitution method, there is generally a relationship in the power of the coefficient - in this case, $4$ - and the power of the denominator in $T()$. Realizing this, you can often quickly generalize and jump to the base condition. To explicitly illustrate:
\begin{align} T(n) &= 4^{1}T(\frac{n}{3^{1}}) + n \\
&= 4^{2}T(\frac{n}{3^{2}}) + 4n + n \\
&= 4^{3}T(\frac{n}{3^{3}}) + 4^{2}n + 4n + n \\
& = \ ... \end{align}
At this point, look to the value within $T()$ in order to determine the remaining number of additional substitutions need to take place. For this problem, the base case is $T(3) = 1$, and,
$$3 = \frac{n}{3^{k-1}}$$
therefore, we get
\begin{align} T(n) &= 4^{k-1}T(\frac{n}{3^{k-1}}) + 4^{k-2}n + ... + 4^{k - (k - 1)}n + n \\ & = 4^{k-1} + 4^{k-2}n + ... + 4^{k - (k - 1)}n + n \end{align}
Now, these terms need to be summed. Re-expressing $4 = 2^{2}$ and factoring out $n$,
\begin{align} T(n) & = 2^{2(k-1)} + 2^{2(k-2)}n + ... + 2^{2(k - [k - 1])}n + n \\
& = 2^{2(k-1)} + n(2^{2(k-2)} + ... + 2^{2(k - [k - 1])} + 1) \end{align}
Recognizing the parenthesized terms collectively represent a geometric series, we can easily sum the terms to get
\begin{align} T(n) &= 2^{2(k-1)} + n\sum_{i=1}^{k-2} \ (2^{2})^{i} \\
&= 2^{2(k-1)} + n(2^{2(k-2)+1} - 1)\end{align}
Lastly, back substitute $k$ to get
\begin{align} T(n) &= 4^{(log_{3}(n)-1)} + n(4^{(log_{3}(n)-2)+1} - 1) \\
&= 4^{log_{3}(n)-1}(n+1)-n \end{align}
Q2: I read that Θ(g(n)) is both the upper bound O(g(n)) and the lower bound Ω(g(n)) so does that mean that the three answers to my question would be...
Using case 1 of the Master Theorem as described by here, $T(n) \in \Theta(n^{1.262})$. Using theta notation alone will suffice for describing the bounds for this function, as it's implied in its definition. Consider the following graph:
Best Answer
It seems the cause of your trouble is simply that you made a mistake while computing $(n/3)^{\log_34}$, which is $n^{\log_34}/4$ and not $n^{\log_34}/3$.
(Note that $(n/3)^{\log_34}=n^{\log_34}/3^{\log_34}$ and that $3^{\log_34}=\exp(\ln3\cdot\log_34)$ with $\ln3\cdot\log_34=\ln3\cdot\ln4/\ln3=\ln4$ hence $3^{\log_34}=4$.)
Anyway, the hint you were given is to assume that $T(n)\leqslant An^{\log_34}+Bn$ $(*)$ and to check if $(*)$ is hereditary for some suitable $B$. Hence, assume $(*)$ holds for $n/3$, then $$ T(n)=4T(n/3)+n\leqslant 4A(n/3)^{\log_34}+4B(n/3)+n=An^{\log_34}+(4B/3+1)n, $$ and one sees that $(*)$ holds for $n$ as soon as $4B/3+1\leqslant B$, for example for $B=-3$.
Now, choosing $A$ large enough such that $T(n)\leqslant An^{\log_34}-3n$ holds for $n$ small, one sees that $T(n)\leqslant An^{\log_34}-3n$ holds for every $n$.
Likewise, there exists $A'$ such that $T(n)\geqslant A'n^{\log_34}-3n$ holds for $n$ small, and this is enough to guarantee that $T(n)\geqslant A'n^{\log_34}-3n$ holds for every $n$. Thus, $T(n)=\Theta(n^{\log_34})$.
In hindsight, all this can be made easier using the change of variable $\bar T(n)=T(n)+3n$ since $\bar T(n)=4T(n/3)+n+3n=4\bar T(n/3)$, a recursion whose solution can be computed directly.