<u>Answer: </u>
<u>For 1:</u> The volume of HCl required is 6 L.
<u>For 2:</u> The volume of HCl required is 9 L.
<u>For 3:</u> The volume of sulfuric acid required is 4.5 L.
<u>Explanation:</u>
To calculate the volume of acid, we use the equation given by neutralization reaction:
 ......(1)
        ......(1)
where,
 are the n-factor, molarity and volume of acid
 are the n-factor, molarity and volume of acid 
 are the n-factor, molarity and volume of base
 are the n-factor, molarity and volume of base
We are given:

Putting values in equation 1, we get:

Hence, the volume of HCl required is 6 L.
We are given:

Putting values in equation 1, we get:

Hence, the volume of HCl required is 9 L.
To calculate the volume of acid, we use the equation:

where,
 are the normality and volume of acid
 are the normality and volume of acid 
 are the normality and volume of base
 are the normality and volume of base
We are given:

Putting values in above equation, we get:

Hence, the volume of sulfuric acid required is 4.5 L.