Solving the recurrence $T(n) = n^{3/4}(^{1/4})+ n $
You can just expand it:
$$
\begin{align*}
T(n) &amp;= n + n^{3/4} T(n^{1/4}) \\ &amp;=
n + n^{3/4}\cdot [n^{1/4}+ n^{3/16}\cdot T(n^{1/16})]  \\ &amp;=
n + n + n^{3/4} \cdot n^{3/16} T(n^{1/16}) \\ &amp;=
n + n + n + n^{3/4} \cdot n^{3/16} \cdot n^{3/64} T(n^{1/64}) \\ &amp;=
\cdots
\end{align*}
$$
We see that $T(n) = C(n) n$, where $C(n)$ is the number of times we need to apply $n \mapsto n^{1/4}$ until the answer drops below some constant.

I'll let you figure out $C(n)$, and so complete this exercise.