problem in understanding mul & imul instructions of Assembly language

There are lots of different variations of the imul instruction.

The variant you've stumbled upon is a 16 bit multiplication. It multiplies the AX register with whatever you pass as the argument to imul and stores the result in DX:AX.

One 32 bit variant works like the 16 bit multiplication but writes the register into EDX:EAX. To use this variant all you have to do is to use a 32 bit source operand.

E.g:

  ; a 16 bit multiplication:
  mov ax, [factor1]
  mov bx, [factor2]
  imul bx              ; 32-bit result in DX:AX
  ; or  imul  word [factor2]

  ; a 32 bit multiplication:
  mov eax, [factor1]
  mov ebx, [factor2] 
  imul ebx             ; 64-bit result in EDX:EAX

---

On a 386 or later, you can also write an imul in the two operand form. That makes it much more flexible and easier to work with. In this variant you can freely choose any 2 registers as the source and destination, and the CPU won't waste time writing a high-half result anywhere. And won't destroy EDX.

  mov   ecx, [factor1]
  imul  ecx, [factor2]    ; result in ecx, no other registers affected
  imul  ecx, ecx          ; and square the result

Or for signed 16-bit inputs to match your imul. (use movzx for unsigned inputs)

  movsx   ecx, word [factor1]
  movsx   eax, word [factor2]  ; sign-extend inputs to 32-bit
  imul    eax, ecx             ; 32-bit multiply, result in EAX
  imul    eax, eax             ; and square the result

This variant of imul was introduced with 386, and is available in 16 and 32-bit operand-size. (And 64-bit operand-size in 64-bit mode).

In 32-bit code you can always assume that 386 instructions like imul reg, reg/mem are available, but you can use it in 16 bit code if you don't care about older CPUs.

186 introduced a 3-operand immediate form.

imul  cx, bx, 123        ; requires 186

imul  ecx, ebx, 123      ; requires 386

Q1/Q2: The x86 instruction set maintains its 16-bit history. When doing a 16-bit multiply, the answer is stored in DX:AX. That's just the way it is, because that's how it was in 16-bit land.

Q3: The code you showed has a bug if you try to compute the square of a number larger than 2^16, because the code ignores the high 32 bits of the result stored in edx.

Q4: I think you may be misreading the table. 8-bit multiplications are stored in a 16-bit result; 16-bit multiplications are stored in a 32-bit result; 32-bit multiplications are stored in a 64-bit result. Which line are you referring to specifically?