That works, but it's doesn't look good.

query.js

r.expr([
  {
    "sub_queries" : [
      {
        "sent_params" : {
          "show_take_out_items" : null,
          "show_in_store_items" : null,
          "show_delivery_items" : null,
          "location_id" : 448,
          "category_id" : 5333
        },
        "rson_query" : "https://www.subtledata.com/API/M/1/?Q1=0205MkxGX0Fa|448|5333|||"
      },
      {
        "sent_params" : {
          "show_take_out_items" : null,
          "show_in_store_items" : null,
          "show_delivery_items" : null,
          "location_id" : 448,
          "category_id" : 5127
        },
        "rson_query" : "&Q2=0205|448|5127|||"
      },
      {
        "sent_params" : {
          "show_take_out_items" : null,
          "show_in_store_items" : null,
          "show_delivery_items" : null,
          "location_id" : 448,
          "category_id" : 5129
        },
        "rson_query" : "&Q3=0205|448|5129|||"
      },
      {
        "sent_params" : {
          "show_take_out_items" : null,
          "show_in_store_items" : null,
          "show_delivery_items" : null,
          "location_id" : 448,
          "category_id" : 5148
        },
        "rson_query" : "&Q14=0205|448|5148|||"
      }
    ]
  }
])
.map(function(doc) {
  return doc.merge({ sub_queries: doc("sub_queries")
    .limit(2)
    .append(doc("sub_queries").nth(2).merge({response: "hello"}))
    .union(doc("sub_queries").skip(3))

  })
})

// Suppose the previous document is in the table temp.
// Since you probably have multiple documents and want to update just one, you have to add a .filter() after .table() and before .update()

r.db('test').table('temp').update(function(doc) {
  return doc.merge({ sub_queries: doc("sub_queries")
    .limit(2)
    .append(doc("sub_queries").nth(2).merge({response: "hello"}))
    .union(doc("sub_queries").skip(3))

  })
})

In python

r.db('test').table('temp').update(
    lambda doc: doc.merge({ 
        "sub_queries": doc["sub_queries"].limit(2)
            .append(doc["sub_queries"].nth(2).merge({"response": "hello"}))
            .union(doc["sub_queries"].skip(3))
    }) 
)

Very helpful! Can you explain the limit/nth/skip pieces? Thanks!

Oh wow.....this worked immediately. Thank you so much!