Anyone explain to (i), (ii)

How can we get the instruction words and R8=?[hex]?

(i) instruction words[hex] is 0x4508, and R8= 0xF002

How can I get that?

(ii) instruction words[hex] is 0x4548 and R8=0x0002

How can I get that?

Consider the following instructions given in the table below. For each instruction determine its length (in words), the instruction words (in hexadecimal), source operand addressing mode, and the content of register R7 after execution of each instruction. Fill in the empty cells in the table. The initial content of memory is given below. Initial value of registers R:5 R6, and R7 is as follows: R5-0xF002, R6-0xF0OA, R8 0xFF88. Assume the starting conditions are the same for each each gmeieon dise always start from intial conditions in memonyz stion (i.e., always start from initial conditions in memory) and given register values. The format of the first word of double-operand instructions is shown below. (Note: Op-code for MOV is 0100b) 5 14 131 1 10 9 8 76543 Op-code S-Reg AdBWAs D-Reg Instr. Address (P Instruction Instr.Instruction Word(s) Source Operand R8-? Addressing Mode Registeir Registeir Length HEX] words HEX] (i)0x1116MOV R5, R8 (ii)0x1116MOV.B R5, R8 (a)Ox1116MOV 4(R5), R8 0x4508 0x4548 0xF002 0x0002 (b)Ox1116MOVB 3(R5), R8 (c)Ox1116MOVB -3(R6), R8 (d)Ox1116MOV TONI, R8 (e)Ox1116MOVB EDE, R8 (f)Ox1116MOV &EDE, R8 (g)Ox1116MOVB @R5, R8 (h)Ox1116MOV @R5+, R8 (i) |0x1116 |MOV #45 R8 (j) | 0x1116 |MOV.B #45 R8