H2SO4 (l) + HI (g) --> H2S (g) + I2 (s)
Method 1:
first check the oxidation no. of different species to know which ones are oxidizing and reducing agents.
then you know sulphuric acid is oxidizing agent while hydrogen iodide is reducing agent.
Reduction:
Step1: H2SO4 + e- > H2S + 4H2O
Stpe2 (balance both the charge and hydrogen): 8H+ + H2SO4 + 8e- > H2S + 4H2O
Oxidation:
2HI > I2 + 2e- + 2H+
Then you get the answer: H2SO4(l) + 8 HI(g) > H2S(g) + 4 I2(s) + 4H2O(l)

Method 2:
The oxidation no. of S decreases from +6 to -2 and that of I increases from -1 to 0.
Then
Step 1: H2SO4 + 8HI > H2S + 4I2
Step 2: H2SO4 + 8HI > H2S + 4I2 + 4H2O


Similiarly two methods can be used to balance BrO3- (aq) + Br- (aq) --> Br2 (aq)  in acidic medium
The oxidation no. of Br in B3O3- decreases from +5 to 0 and that of Br in Br- increases from -1 to 0.
Step 1: BrO3- + 5Br- > 3Br2
Step 2 (balance oxygen) :
BrO3- + 5 Br- > 3Br2 + 3H2O
Step 3( balance charge and hydrogen by adding H+ to the left which suits the condition in the question):
6H+(aq) + BrO3-(aq) + 5Br-(aq) > 3Br2(aq) + 3H2O(l)

Try to add sth that can serve multi-purposes such as balancing both the charge and some particles.
Of course you should pay attention to the condition provided in the question, which usually also serves as a hint.

Method 2:
The oxidation no. of S decreases from +6 to -2 and that of I increases from -1 to 0.
Then
Step 1: H2SO4 + 8HI > H2S + 4I2
Step 2: H2SO4 + 8HI > H2S + 4I2 + 4H2O
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It is reminded that it is just the oxidation number that decreases but not the no. of electron.
To balance the equation, i figure out it this way:
the total oxidation number changed is 8*-1 - 4*0=-8
multiplying both sides by two will not balance the equation since the I is not balanced, i.e. 8HI...8I2<-strange!