How to get the sum, difference, product, and quotient from a macro in ConTeXt or Plain TeX?

You can use the ε-TeX primitive \numexpr for expandable integer expressions. The only restriction is that it won't work in engines without ε-TeX extensions (mostly Knuth TeX nowadays), in which case you need to go with siracusa's solution.

These macros being expandable, allows you to use them anywhere TeX expects a number, like in \ifnum, \ifcase, or \ifdim (with the proper units) tests, register assignments, etc.

Here are the four interfaces you want (I swapped the order of the arguments in \dividenumbers because I think it makes more sense this way), plus a generic \inteval, which takes an integer expression, like 2+5*(3-4), as argument:

\def\addnumbers#1#2{\number\numexpr#1+#2\relax}
\def\subtractnumbers#1#2{\number\numexpr#1-#2\relax}
\def\multiplynumbers#1#2{\number\numexpr#1*#2\relax}
\def\dividenumbers#1#2{\number\numexpr#1/#2\relax}
\def\inteval#1{\number\numexpr#1\relax}

\def\test#1#2#3#4{$#2#3#4 = #1{#2}{#4}$\par}
\test\addnumbers{2}+{5}
\test\subtractnumbers{3}-{5}
\test\multiplynumbers{2}\times{-2}
\test\dividenumbers{8}\div{2}

$2+5\times(3-4) = \inteval{2+5*(3-4)}$
\bye

If you don't want to restrict yourself to integers, you can use expl3's FPU, which is format-independent, and also expandable. It allows you a much larger range of operations. The code below defines the four interfaces you asked, plus \fpeval for generic floating point expressions (in LaTeX the package xfp provides the same \fpeval):

\input expl3-generic.tex
\ExplSyntaxOn
\cs_new:Npn \addnumbers #1 #2
  { \fp_eval:n { #1 + #2 } }
\cs_new:Npn \subtractnumbers #1 #2
  { \fp_eval:n { #1 - #2 } }
\cs_new:Npn \multiplynumbers #1 #2
  { \fp_eval:n { #1 * #2 } }
\cs_new:Npn \dividenumbers #1 #2
  { \fp_eval:n { #1 / #2 } }
\cs_new:Npn \fpeval #1
  { \fp_eval:n { #1 } }
\ExplSyntaxOff
\def\test#1#2#3#4{$#2#3#4 = #1{#2}{#4}$\par}
%
\test\addnumbers{2}+{5}
\test\subtractnumbers{3}-{5}
\test\multiplynumbers{2}\times{-2}
\test\dividenumbers{8}\div{2}

$2^2+5\times(\cos(3)-4) = \fpeval{2^2+5*(cos(3)-4)}$
\bye

As you haven't specified if the solution should be expandable, here's a non-expandable plain TeX solution.

We first define a general \docalc macro which takes a calculation command working on count registers (\advance, \multiply or \divide), and the two operands. The operation is applied on a local register which value is immediately output afterwards. Based on \docalc we then define the four macros for doing the basic arithmetic operations:

\def\docalc#1#2#3{%
    \begingroup
    \count0=#3
    #1\count0 by #2\relax
    \the\count0
    \endgroup
}

\def\addnumbers       {\docalc\advance}
\def\subtractnumbers#1{\docalc\advance{-#1}}
\def\multiplynumbers  {\docalc\multiply}
\def\dividenumbers    {\docalc\divide}

$2 + 5 = \addnumbers{2}{5}$

$5 - 3 = \subtractnumbers{3}{5}$

$2 \cdot (-2) = \multiplynumbers{2}{-2}$

$8 \div 2 = \dividenumbers{2}{8}$
\bye

In ConTeXt you can use the lua interface:

\starttext
  The first will display the number \ctxlua{context(2+5)}
\stoptext