(a) Write and run a program using 8086 assembly language that increments a byte value stored in a memory location by a value 2. The result should be stored in the same memory location. For example, if a memory location contains 0101 0001, then the program will add 2 to this value and store the result 0101 0011 (after adding 2) in the same location. 20

 Certainly! Below is a simple 8086 assembly language program that increments a byte value stored in a memory location by 2. The result is stored back in the same memory location:

```assembly

section .data

    value db 51h  ; Initial byte value, change as needed

section .text

    global _start

_start:

    mov al, [value]  ; Load the byte value into AL register

    add al, 2         ; Increment AL by 2

    mov [value], al   ; Store the result back in the memory location

    ; Add any additional code here if needed

    ; Exit the program

    mov eax, 1        ; System call number for sys_exit

    xor ebx, ebx      ; Exit code 0

    int 0x80          ; Call kernel

```

This program uses Linux system calls for simplicity (`int 0x80`). If you are running it on a modern 64-bit system, you might need to adjust it accordingly.

Compile and run the program using an assembler and linker. For example, using NASM and LD:

```bash

nasm -f elf32 your_program.asm -o your_program.o

ld -m elf_i386 -s -o your_program your_program.o

./your_program

```

Make sure to replace `your_program.asm` with the actual filename you save the program in. The program increments the initial byte value (`51h`) by 2, and the final value is stored back in the same memory location.