Here is a more formal approach to this problem, and to similar ones. Let's write $\;T(x)\;$ for "$\;x\;$ speaks the Truth" or "$\;x\;$ is a knight". The key insight is that if individual $\;x\;$ says $\;P\;$, then $\;T(x) \equiv P\;$.
$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Tag}[1]{\text{(#1)}}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
\newcommand{\exactly}[4]{\text{exactly one of }#1,#2,#3\text{ is }#4}
$Using this insight, we can formalize the statements of these individuals as follows:
\begin{align}
\tag 1 T(N1) &\;\equiv\; T(N1) \land T(N2) \land T(N3) \\
\tag 2 T(N2) &\;\equiv\; \exactly{T(N1)}{T(N2)}{T(N3)}{\true} \\
\tag 3 T(N3) &\;\equiv\; T(N3) \land \lnot T(N1) \land \lnot T(N2) \\
\end{align}
First, we can substitute both $\Tag 1$ and $\Tag 3$:
$$\calc
T(N1)
\calcop\equiv{using $\Tag 1$}
T(N1) \land T(N2) \land T(N3)
\calcop\equiv{using $\Tag 3$}
T(N1) \land T(N2) \land T(N3) \land \lnot T(N1) \land \lnot T(N2)
\calcop\equiv{contradiction (two, actually): $P \land \lnot P \;\equiv\; \false$}
\false
\endcalc$$
In other words, $N1$ is a knave. Then, we can use this to simplify $\Tag 2$:
$$\calc
\tag 2
T(N2) \;\equiv\; \exactly{T(N1)}{T(N2)}{T(N3)}{\true}
\calcop\equiv{simplify using $\;T(N1) \equiv \false\;$}
T(N2) \;\equiv\; T(N2) \not\equiv T(N3)
\calcop\equiv{simplify using $\;P \equiv P \;\equiv\; \true\;$}
T(N3) \;\equiv\; \false
\endcalc$$
So $N3$ is also a knave.
There is nothing we can say about $N2$: given what we know, it could be either knight or knave. That is, substituting $\;T(N1) \equiv \false\;$ and $\;T(N3) \equiv \false\;$ in $\Tag 1, \Tag 2, \Tag 3$ makes all three true, regardless of $\;T(N2)\;$.