Remainder of $3^n$ modulo $15$ Let $n \in \mathbb{N}_{>0}$. Then we have $$\begin{align*} &3 \equiv 3 \bmod 15\\&3^2 \equiv 9 \bmod 15\\&3^3 \equiv 12 \bmod 15\\&3^4 \equiv 6 \bmod 15\\&3^5 \equiv 3 \bmod 15\end{align*}$$ Hence all remainders modulo $15$ of $3^n$ should be $3,9,12$ and $6$. Why is this the case? Also, How could I calculate $$3^{1291}\equiv ? \bmod 15$$

The only choices are $3, 6, 9, 12$ because $15$ is divisible by $3$, so what you're really doing is calculating $3^{n-1} \bmod 5$ and multiplying by $3$.  That can't be $0$ (powers of $3$ can't be multiples of $5$ by unique factorization), so the only choices are $1, 2, 3, 4$ multiplied by $3$.
As for your second problem, notice that the pattern repeats: $3, 9, 12, 6, 3, 9, 12, 6, 3, 9, 12, 6, \ldots$  So take $u = 1291$ modulo ____ (fill in the blank) and then $3^u \equiv 3^{1291} \bmod 15$.

Oh, in case the reasoning isn't clear, we use the fact that
$$
ab \bmod 3 = (a \bmod 3)(b \bmod 3) \bmod 3
$$
In other words, we can take the modulo before we multiply, making things a lot less computationally intensive.
