Q.Try the following conversions.

Work each by place value (to decimal) or digit-grouping (octal to binary):

(i)   (514)8  = 5x64 + 1x8 + 4x1        = 320+8+4          = (332)10
(ii)  (220)8  : each octal digit -> 3 bits: 2=010, 2=010, 0=000
              -> 010 010 000            = (10010000)2
(iii) (76F)16 = 7x256 + 6x16 + 15x1     = 1792+96+15       = (1903)10
(iv)  (4D9)16 = 4x256 + 13x16 + 9x1     = 1024+208+9       = (1241)10
(v)   (11001010)2 = 128+64+0+0+8+0+2+0  = (202)10
(vi)  (1010111)2  = 64+0+16+0+4+2+1     = (87)10

The idea. A number written in base r means: each digit times r raised to its position (position 0 at the right). So converting to decimal is just "multiply and add". Converting octal to binary is even easier — since 8 = 2x2x2, each octal digit expands independently into exactly 3 binary bits.

(i) (514)8 to decimal

(514)8 = 5x8^2 + 1x8^1 + 4x8^0
       = 5x64  + 1x8   + 4x1
       = 320 + 8 + 4
       = (332)10

(ii) (220)8 to binary — replace each octal digit by its 3-bit group:

2 -> 010    2 -> 010    0 -> 000
(220)8 = 010 010 000 = (10010000)2   (leading zero dropped)
Check: (10010000)2 = 128 + 16 = 144 = 2x64 + 2x8 + 0 = (220)8  correct

(iii) (76F)16 to decimal — remember F = 15:

(76F)16 = 7x16^2 + 6x16^1 + 15x16^0
        = 7x256  + 6x16   + 15x1
        = 1792 + 96 + 15
        = (1903)10

(iv) (4D9)16 to decimal — D = 13:

(4D9)16 = 4x256 + 13x16 + 9x1
        = 1024 + 208 + 9
        = (1241)10

(v) (11001010)2 to decimal — place values 128, 64, 32, 16, 8, 4, 2, 1:

(11001010)2 = 1x128 + 1x64 + 0x32 + 0x16 + 1x8 + 0x4 + 1x2 + 0x1
            = 128 + 64 + 8 + 2
            = (202)10

(vi) (1010111)2 to decimal:

(1010111)2 = 1x64 + 0x32 + 1x16 + 0x8 + 1x4 + 1x2 + 1x1
           = 64 + 16 + 4 + 2 + 1
           = (87)10

Verify all six with Python:

print(int('514', 8))          # octal string to decimal
print(bin(int('220', 8)))     # octal to binary
print(int('76F', 16))
print(int('4D9', 16))
print(int('11001010', 2))
print(int('1010111', 2))

332
0b10010000
1903
1241
202
87