Answer :
The formula for the pH is the following:
[tex]pH=-\log _{10}\lbrack H^+\rbrack[/tex]Where the concentration is in mol L⁻¹.
So, if we want the formula to calculate the concentration given the pH, we have to invert it:
[tex]\begin{gathered} pH=-\log _{10}\lbrack H^+\rbrack \\ -\log _{10}\lbrack H^+\rbrack=pH \\ \log _{10}\lbrack H^+\rbrack=-pH \\ \lbrack H^+\rbrack=10^{-pH} \end{gathered}[/tex]Now, let's answer the items:
(a) It turns red for pH 4.4 or lower. So:
[tex]\begin{gathered} \lbrack H^+\rbrack=10^{-pH} \\ \lbrack H^+\rbrack=10^{-4.4} \\ \lbrack H^+\rbrack\approx3.98\cdot10^{-5}mol\, L^{-1} \end{gathered}[/tex]The methyl red turn red for a concentration of H⁺ of approximately 3.98 10⁻⁵mol L⁻¹ or higher.
(b) It turns yellow for pH 6.2 or higher. so:
[tex]\begin{gathered} \lbrack H^+\rbrack=10^{-pH} \\ \lbrack H^+\rbrack=10^{-6.2} \\ \lbrack H^+\rbrack\approx6.31\cdot10^{-7}mol\, L^{-1} \end{gathered}[/tex]The methyl red turn yellow for a concentration of H⁺ of approximately 6.31 10⁻⁷mol L⁻¹ or lower.
(c) In a concentration of H⁺ of 3.2 10⁻⁴mol L⁻¹, we have:
[tex]\begin{gathered} pH=-\log _{10}\lbrack H^+\rbrack \\ pH=-\log _{10}(3.2\times10^{-4}) \\ pH\approx-(-3.5) \\ pH\approx3.5 \end{gathered}[/tex]Which is lower then 4.4, so it will turn red.