\ifthenelse and modulo

You can use pgfmath facilities or l3fp.

\documentclass{article}
\usepackage{tikz,ifthen}

\begin{document}

\begin{tikzpicture}
  \foreach \i in {1,...,12}{
    \pgfmathsetmacro\test{mod(\i,3)==0}
    \ifthenelse {\test=1}
      {\draw[purple] (.75*\i,0) node ()  {\i};}
      {\draw[blue] (.75*\i,0) node ()  {\i};}
  }
\end{tikzpicture}

\begin{tikzpicture}
  \foreach \i in {1,...,12}{
    \edef\test{\fpeval{\i/3==trunc(\i/3)}}
    \ifthenelse {\test=1}
      {\draw[purple] (.75*\i,0) node ()  {\i};}
      {\draw[blue] (.75*\i,0) node ()  {\i};}
  }
\end{tikzpicture}

\end{document}

Or you can use a reimplementation of \ifthenelse.

\documentclass{article}
\usepackage{tikz}

% https://tex.stackexchange.com/a/467527/4427
\ExplSyntaxOn
\NewExpandableDocumentCommand{\xifthenelse}{mmm}
 {
  \bool_if:nTF { #1 } { #2 } { #3 }
 }

\cs_new_eq:NN \numtest     \int_compare_p:n
\cs_new_eq:NN \oddtest     \int_if_odd_p:n
\cs_new_eq:NN \fptest      \fp_compare_p:n
\cs_new_eq:NN \dimtest     \dim_compare_p:n
\cs_new_eq:NN \deftest     \cs_if_exist_p:N
\cs_new_eq:NN \namedeftest \cs_if_exist_p:c
\cs_new_eq:NN \eqdeftest   \token_if_eq_meaning_p:NN
\cs_new_eq:NN \streqtest   \str_if_eq_p:ee
\cs_new_eq:NN \emptytest   \tl_if_empty_p:n
\cs_new_eq:NN \blanktest   \tl_if_blank_p:n
\cs_new_eq:NN \booleantest \legacy_if_p:n
\cs_new:Npn \modetest #1
 {
  \str_case:nnF { #1 }
   {
    {h}{\mode_if_horizontal_p:}
    {v}{\mode_if_vertical_p:}
    {m}{\mode_if_math_p:}
    {i}{\mode_if_inner_p:}
   }
   {\c_false_bool}
 }
\cs_new:Npn \enginetest #1
 {
  \str_case:nnF { #1 }
   {
    {luatex}{\sys_if_engine_luatex_p:}
    {pdftex}{\sys_if_engine_pdftex_p:}
    {ptex}{\sys_if_engine_ptex_p:}
    {uptex}{\sys_if_engine_uptex_p:}
    {xetex}{\sys_if_engine_xetex_p:}
   }
   {\c_false_bool}
 }
\ExplSyntaxOff


\begin{document}

\begin{tikzpicture}
  \foreach \i in {1,...,12}{
    \xifthenelse {\fptest{\i/3==trunc(\i/3)}}
      {\draw[purple] (.75*\i,0) node ()  {\i};}
      {\draw[blue] (.75*\i,0) node ()  {\i};}   
  }
\end{tikzpicture}   

\end{document}

For the \xifthenelse version, you can also add a user level function for the modulo operation: add

\cs_new_eq:NN \IntMod \int_mod:nn

just before \ExplSyntaxOff and the test can become

\xifthenelse{\numtest{\IntMod{\i}{3}==0}}

Consider using \fpeval{} rather than pgfmath if you do not need compatibility with old releases of LaTeX. This is faster, more accurate and significantly less likely to result in the too-large errors which are the bane of pgfmath users' existence.

\fpeval{} is a 2e interface to the l3fp programming module. Although it slightly oddly lacks a mod function, it does have trunc, which provides a reasonable alternative.

As far as I know, there is no 2e interface to the programming layer's if <compare fps> then <something> else <something else>, so \ifnum remains the best option here, but we can use \fpeval to compare the expressions and insert 1 or 0 into the input stream. Then we only need \ifnum<result>=1 ... \else ... \fi, which isn't at risk of exceeding any TeX limitation.

\documentclass{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
  \foreach \i in {1,...,12}{
    \ifnum\fpeval{(\i/3)=(trunc(\i/3)) ? 1 : 0}=1
      \draw[purple] (.75*\i,0) node ()  {\i};
    \else
      \draw[blue] (.75*\i,0) node ()  {\i};
    \fi
  }
\end{tikzpicture}   
\end{document}

You can use \ifnum TeX primitive, no \ifthenelse macro is needed. Your example looks like this:

\begin{tikzpicture}
  \foreach \i in {1,...,12}{%
    \ifnum \i=\numexpr\i/3*3\relax
      \draw[purple] (.75*\i,0) node {\i};\else
      \draw[blue]   (.75*\i,0) node {\i};\fi
  }
\end{tikzpicture}