Answer by J_H for Calculator final project in Python

    elif choice == '2':
        instructions()
        main()

Ummm, that's probably not exactly what you want. Python lacks a goto statement, so the usual idiom would be to wrap the whole thing in some sort of while loop, perhaps while True. Consider an input sequence of "2 2 2 2 2", or "z z z z z". You'll wind up pushing main onto the call stack again, and again, and again....

Your locals never go out of scope, so their ref count never decrements and they won't be deallocated and tidied up.

tl;dr: leaking stack frames is Bad, don't do it.

            print("Last Result: "+str(result))

Usual idiom would be print('Last Result:', result), but no harm done.

                num1 = int(input("First Number: "))

The requirements seemed pretty clear about "...must work with decimal numbers." You chose to invoke int() rather than float().

            if result != 0:     
                num1 = result

I can't imagine how that corresponds to The Right Thing. If you want a sentinel, None would be much better than 0, just consider a sequence like 2 + 1 = - 3 = (which yields zero). Perhaps there's a reason for assigning the accumulator to num1, but I'm not seeing it. It appears you have discarded the user input.

            operator = input("Opertaor: ").lower()

Typo.

                calculator()

Same criticism as above. Python lacks goto, and you're leaking stack frames here.

            if operator == "sqrt":
                result = (sqrt(num1))
                print("Result: "+str(result))
                continue

It is apparent that I'm not understanding why you are using the result accumulator versus the num1 user input. On most calculators, repeated clicking of the √ key would yield repeated assignment of result = sqrt(result). I'm a little concerned about num1 versus result confusion.

            if operator=="esc":
                main()

Same remark about leaking stack frames. A return statement would be usual, here.

            if operator.isdigit(): 

This seems a bit restrictive. Better to unconditionally raise an "unrecognized operator" error.

            num2 = int(input("Second Number: "))

Same criticism as above, the "decimal" instructions are pretty clear about calling float().

                result = (add(num1,num2))

All four of these assignments seem pretty weird, as for a typical calculator we'd be accumulating result = add(result, num2). This is a consequence of the num1 assignment above. Also, please discard the (extraneous parentheses).

def sqrt(num1):
    return (num1**(1/2.0))

This works fine, but do consider the usual idiom of simply calling math.sqrt().