What are the oxidation number of the underlined elements

(a) KI₃
let the oxidation number of I is x
+1 + 3x = 0
    x = -1/3
hence, the average oxidation number of I is -1/3 .but we know, oxidation number can't be fractional. actually,
       the structure of KI₃,     K⁺ (I⁰------I⁰<----I⁻)⁻
in this structure, a co-ordination bond is formed between I₂ molecule and I⁻ ion. hence, oxidation of I₃ is -1

(b)H₂S₄O₆
let the oxidation number of S = x
 +2+4x +6(-2)=0
  x = +5/2 
similarly, oxidation number of this compound depends upon structure of molecule. 
see structure,      O                       O
                              ||                        ||
          H---O---------S⁺⁵----S⁰-----S⁰-------S⁺⁵--------O-----H
                              ||                        || 
                              O                      O
after seeing the structure, it is clear that oxidation number of Satoms  are +5,0,0, +5 respectively.

(c)Fe₃O₄
let the oxidation number of Fe is x 
then, 3x + 4(-2) = 0
    x = + 8/3 
 hence, the average oxidation number of Fe is +8/3, but stoichiometrically, Fe₃O₄ is an equimolar mixture of FeO and Fe₂O₃. 
                 The oxidation number of Fe in FeO is +2 
and the oxidation number of Fe in Fe₂O₃ is +3 
then, 2x + x = +6 + 2 = +8 ⇒ x = +8/3
(d) CH₃CH₂OH
let the o.n of C is x 
  x + 3 + x + 2 - 2 + 1 =0
      x = -2 
therefore, average o.n of C is -2 
now, see the structure of given molecule.
              H                H
               |                  |
  H------- C² ------------- C¹ ---- O----H   [CH₃⁺ and CH₂OH⁻
              |                   |
             H                  H
    now, oxidation number of C₁ atom : x+ 2 -2 +1 = -1 ⇒ x = -2
    oxidation number of C₂ atom : x + 3 = +1 ⇒ x = - 2
thats why average oxidation number of C - atoms is -2 

(e) CH₃COOH
     let the oxidation number of C is x 
x + 3 + x -2 -2 + 1 = 0
  2x = 0  ⇒x = 0
therefore, average oxidation number of C is zero. 
                  let see the structure for understanding 
          H           O
           |            ||
H------C²----------C¹------O---H    [CH₃⁺ and COOH⁻ 
          |
         H
                now, oxidation number of C₁ atom : x - 2 -2 +1 =1 ⇒ x = +2
                        oxidation number of C₂ atom : x +3 = +1 ⇒ x = -2 
that's why average oxidation number of CH₃COOH is zero.