How to Detect a Pagebreak?

Questions
5

Because it might be useful, here’s a different example but still using the repeat/count mechanism, taken from discord

#set page(margin: 1cm, width: 10cm, height: 6cm)
#let rep-count-header = counter("my-table-header-repeat")
#let repheader(first, cont) = {
  table.header(table.cell({
    rep-count-header.step()
    context {
      if rep-count-header.get().first() == 1 {
        first
      } else {
        cont
      }
    }
  }))
}

#let continuation-footer(cont, stroke: none) = {
  table.footer({
    table.cell(inset: 0pt, stroke: none, {
      rep-count-header.step()
      context {
      // show continuation footer if we have header continuations
      // and we're not on the last one.
      let s = rep-count-header.final()
      let t = rep-count-header.get()
      if t.first() < s.first() {
        set align(right)
        set block(stroke: (top: none, rest: stroke))
        show: pad.with(0.45em)  // manual inset
        cont
      } else {
        // Need to make the cell "invisible" / take no space when not used
        none
      }
    }})
  })
}

#show math.equation: set block(breakable: true)

#table(
  stroke: 0.2pt,
  repheader([Example], [Example (cont'd)]),
  {
  $
    [ sqrt(2)(cos pi/8 + i sin pi/8)]^12: \
    [ sqrt(2)(cos pi/8 + i sin pi/8)]^12 = \
    // just filler here
    = [ sqrt(2)(cos pi/8 + i sin pi/8)]^12 \
    = [ sqrt(2)(cos pi/8 + i sin pi/8)]^12 \
    = [ sqrt(2)(cos pi/8 + i sin pi/8)]^12 \
  $
  },
  continuation-footer(stroke: 0.2pt)[continued $->$]
)

Just because it might be simpler to extract the parts you need from a simpler example.