1)  a.     \(T\left(n\right)\ =\ 3T\left(\frac{n}{4}\right)\ +\ \sqrt{n},\ T\left(1\right)\ =\ 1\)
                   Applying Master's Theorem, we get:
                    \(a\ =\ 3,\ b\ =4\)
                \(\Rightarrow n^{\log_34}\ and\ f\left(n\right)=\sqrt{n}\)
                    \(\log_34\ \approx0.79\ \Rightarrow n^{0.79}>\ f\left(n\right)\)
                    \(T\left(n\right)\ =\ \theta\left(n^{0.79}\right);\ \ f\left(n\right)\ is\ smaller\ by\ a\ factor\ of\ \epsilon\ \approx\ 0.3\) 
b.        \(T\left(n\right)=9T\left(\frac{n}{3}\right)+5n^2,\ T\left(1\right)\ =\ 1\)
        Applying Master's Theorem, we get:
            \(a=9,\ b=3\)
            \(n^{\log_39}=n^2\)
            \(f\left(n\right)=5n^2\)
          \(\ \Rightarrow T\left(n\right)\ =\ \theta\left(n^2\log n\right)\) [case 2]
c.         \(T\left(n\right)=T\left(\frac{n}{3}\right)+T\left(\frac{2n}{3}\right)+cn^2\)
            After expanding the recursion tree, we can see the cost at \(i\) depth is  \(c\left(\frac{4}{9}\right)^in^2\)