You can use a Natural Language Tagger, which would work even if both scripts are intermingled:

import NaturalLanguage

let str = "¿como? بداية start وسط middle начать средний конец نهاية end. 從中間開始. "

let tagger = NLTagger(tagSchemes: [.script])

tagger.string = str

var index = str.startIndex
var dictionary = [String: String]()
var lastScript = "other"


while index < str.endIndex {
    let res = tagger.tag(at: index, unit: .word, scheme: .script)
    let range = res.1

    let script = res.0?.rawValue

    switch script {
    case .some(let s):
        lastScript = s
        dictionary[s, default: ""] += dictionary["other", default: ""] + str[range]
        dictionary.removeValue(forKey: "other")
    default:
        dictionary[lastScript, default: ""] += str[range]
    }

    index = range.upperBound
}

print(dictionary)

and print the result if you'd like:

for entry in dictionary {
    print(entry.key, ":", entry.value)
}

yielding :

Hant : 從中間開始. 
Cyrl : начать средний конец 
Arab : بداية وسط نهاية 
Latn : ¿como? start middle end.

This is still not perfect since the language tagger only checks to which script the most number of letters in a word belong to. For example, in the string you're working with, the tagger would consider الصَّالِحِينَ.Bismika as one word. To overcome this, we could use two pointers and traverse the original string and check the script of words individually. Words are defined as contiguous letters:

let str = "بِاسْمِكَ رَبِّي وَضَعْتُ جَنْبِي، وَبِكَ أَرْفَعُهُ، فَإِنْ أَمْسَكْتَ نَفْسِي فَارْحَمْهَا، وَإِنْ أَرْسَلْتَهَا فَاحْفَظْهَا، بِمَا تَحْفَظُ بِهِ عِبَادَكَ الصَّالِحِينَ.Bismika rabbee wadaAAtu janbee wabika arfaAAuh, fa-in amsakta nafsee farhamha, wa-in arsaltaha fahfathha bima tahfathu bihi AAibadakas-saliheen. In Your name my Lord, I lie down and in Your name I rise, so if You should take my soul then have mercy upon it, and if You should return my soul then protect it in the manner You do so with Your righteous servants."

let tagger = NLTagger(tagSchemes: [.script])
var i = str.startIndex
var dictionary = [String: String]()
var lastScript = "glyphs"

while i < str.endIndex {
    var j = i
    while j < str.endIndex,
        CharacterSet.letters.inverted.isSuperset(of: CharacterSet(charactersIn: String(str[j]))) {
        j = str.index(after: j)
    }
    if i != j { dictionary[lastScript, default: ""] += str[i..<j] }
    if j < str.endIndex { i = j } else { break }

    while j < str.endIndex,
        CharacterSet.letters.isSuperset(of: CharacterSet(charactersIn: String(str[j]))) {
        j = str.index(after: j)
    }

    let tempo = String(str[i..<j])
    tagger.string = tempo
    let res = tagger.tag(at: tempo.startIndex, unit: .word, scheme: .script)

    if let s = res.0?.rawValue {
        lastScript = s
        dictionary[s, default: ""] += dictionary["glyphs", default: ""] + tempo
        dictionary.removeValue(forKey: "glyphs")
    }
    else { dictionary["other", default: ""] += tempo }

    i = j
}

You can use the NaturalLanguageTagger as answered by @ielyamani but the only limitation is that it is iOS 12+

If you are trying to do this on earlier iOS versions, you can take a look at NSCharacterSet

You can create your own characterset to check whether a string has english characters and numbers

extension String {

     func containsLatinCharacters() -> Bool {

        var charSet = NSCharacterSet(charactersInString: "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890")
        charSet = charSet.invertedSet

        let range = (self as NSString).rangeOfCharacterFromSet(charSet)

        if range.location != NSNotFound {
            return false
        }

        return true
    }
}

Another option is to use the charactersets already available:

let nonLatinString = string.trimmingCharacters(in: .alphanumerics)//symbols will still get through
let latinString = string.trimmingCharacters(in: CharacterSet.alphanumerics.inverted)//symbols and non-latin characters wont get through

With these you can get the strings you want quite easily. But if these are not good enough, you can look to create your own characterset, use union, intersect etc to filter out the wanted and the unwanted characters.

Step 1: You have to split whole string into an array by "." as I can see there are "." between sentence.

Step 2: Pass each sentence to determine its language and append into different string.

Final Code

//add in your viewController

enum Language : String {
case arabic = "ar"
case english = "en"
}

override func viewDidLoad() {
    super.viewDidLoad()
    //make array of string
    let kalmaArray = "بِاسْمِكَ رَبِّي وَضَعْتُ جَنْبِي، وَبِكَ أَرْفَعُهُ، فَإِنْ أَمْسَكْتَ نَفْسِي فَارْحَمْهَا، وَإِنْ أَرْسَلْتَهَا فَاحْفَظْهَا، بِمَا تَحْفَظُ بِهِ عِبَادَكَ الصَّالِحِينَ.Bismika rabbee wadaAAtu janbee wabika arfaAAuh, fa-in amsakta nafsee farhamha, wa-in arsaltaha fahfathha bima tahfathu bihi AAibadakas-saliheen. In Your name my Lord, I lie down and in Your name I rise, so if You should take my soul then have mercy upon it, and if You should return my soul then protect it in the manner You do so with Your righteous servants.".components(separatedBy: ".")

    splitInLanguages(kalmaArray: kalmaArray)

}



private func splitInLanguages(kalmaArray: [String]){
    var englishText = ""
    var arabicText = ""

    for kalma in kalmaArray {

        if kalma.count > 0 {

            if let language = NSLinguisticTagger.dominantLanguage(for: kalma) {
                switch language {
                case Language.arabic.rawValue:
                    arabicText.append(kalma)
                    arabicText.append(".")
                    break
                default: // English
                    englishText.append(kalma)
                    englishText.append(".")
                    break
                }
            } else {
                print("Unknown language")
            }
        }
    }

    debugPrint("Arabic: ", arabicText)
    debugPrint("English: ", englishText)
}

I hope it will help you to split the string in two language. Let me know if you are still having any issue.