$$\def\t#1{T\left (n/3^{#1}\right )}
\begin{align}
T(n) &= 5\cdot \t1 + 7\\[1em]
\hdashline
&= 5 \times \Biggl (5\cdot \t2 + 7 \Biggr ) + 7\\[1em]
&= 5^2 \cdot \t2 + 5^1 \cdot 7 + 7\\[1em]
\hdashline
&= 5^2 \times \Biggl (5\cdot \t3 + 7 \Biggr ) + 5^1 + 7\\[1em]
&= 5^3 \cdot \t3 + 5^2 \cdot 7 + 5^1 \cdot 7 + 7\\
\hdashline\\[1em]
&\vdots\\
&= 5^k\cdot\t k + 7 \cdot \sum_{i=0}^{k-1} 5^i
\end{align}$$
When $k = \log_3 n$, we have:
$$\begin{align}
T(n) &= 5^{\log_3 n}\cdot T(1) + 7 \cdot \sum_{i=0}^{(\log_3 n)-1} 5^i\\[1em]
&= 5^{\log_3 n}\cdot T(1) + 7 \times \left ( \frac{5^{1 + (\log_3 n - 1)} - 1}{5 - 1} \right )
\end{align}$$
It is given that $T(1) = 5$, hence, we get:
$$\begin{align}
T(n) &= \frac1 4 \cdot \Bigl ( (4\cdot 5 + 7) \cdot 5^{\log_3 n} - 7\Bigr )\\[1em]
&= \frac1 4 \cdot \Bigl(27 \cdot n^{\log_3 5} - 7 \Bigr )\\[1em]
&\lessapprox \frac1 4 \cdot \Bigl (27 \cdot n^{1.465} - 7 \Bigr )
\end{align}$$