Given a Poisson process $N(t)$ with rate $2$, find $P( N(3) = 4 | N(1) = 1 )$. 
Given a Poisson process $N(t)$ with rate $2$, find
  a. $P( N(3) = 4 | N(1) = 1 )$
  b. $P( N(1) = 1 | N(3) = 4 )$    

Using the method, I have 
$$\frac{P( N(3) = 4 \cap N(1) = 1 )}{P( N(1) = 1)}$$
$$P( N(1) = 1) = 2 / \exp(-2)$$
Then because $P( N(3) = 4 \cap N(1) = 1 )$ is independent, therefore
$$\frac{P( N(3) = 4 ) P( N(1) = 1 )}{P( N(1) = 1)} = P( N(3) = 4 )$$
Am I correct in saying this for the first a?

That's false, they are not independent.
Notice that
$$\{N(3)=  4, N(1) = 1\}\iff\{N(1) = 1,N(3)-N(1) = 3\}$$
and 
$N(1)$ is independent of $N(3)-N(1)$. Hence
\begin{align*}
P(N(3) = 4|N(1) = 1) &= \frac{P(N(3) = 4,N(1) = 1)}{P(1) = 1)}\\
&=\frac{P(N(3)-N(1) = 3, N(1) = 1)}{P(N(1) = 1)}\\
&=\frac{P(N(3)-N(1) = 3)P(N(1) = 1)}{P(N(1) = 1)}\\
&=P(N(3)-N(1) = 3)
\end{align*}
