To find the coefficient of \(x\) in the expansion of \(\left( \frac{1}{x} + 3x^2 \right)^5\), we use the binomial theorem:

\(\sum_{k=0}^{5} \binom{5}{k} \left( \frac{1}{x} \right)^{5-k} (3x^2)^k\)

We need the term where the power of \(x\) is 1. The general term is:

\(\binom{5}{k} \left( \frac{1}{x} \right)^{5-k} (3x^2)^k = \binom{5}{k} \cdot \frac{1}{x^{5-k}} \cdot 3^k \cdot x^{2k}\)

This simplifies to:

\(\binom{5}{k} \cdot 3^k \cdot x^{2k - (5-k)} = \binom{5}{k} \cdot 3^k \cdot x^{3k - 5}\)

We set the exponent of \(x\) to 1:

\(3k - 5 = 1\)

Solving for \(k\):

\(3k = 6\)

\(k = 2\)

Substitute \(k = 2\) back into the expression for the coefficient:

\(\binom{5}{2} \cdot 3^2 = 10 \cdot 9 = 90\)

Thus, the coefficient of \(x\) is 90.

To find the coefficient of \(x\) in the expansion of \(\left( \frac{x}{3} + \frac{9}{x^2} \right)^7\), we use the binomial theorem.

The general term in the expansion is given by:

\(\binom{7}{r} \left( \frac{x}{3} \right)^{7-r} \left( \frac{9}{x^2} \right)^r\)

Simplifying, we have:

\(\binom{7}{r} \left( \frac{x^{7-r}}{3^{7-r}} \right) \left( \frac{9^r}{x^{2r}} \right) = \binom{7}{r} \frac{9^r}{3^{7-r}} x^{7-r-2r}\)

We need the power of \(x\) to be 1, so:

\(7 - r - 2r = 1\)

\(7 - 3r = 1\)

\(3r = 6\)

\(r = 2\)

Substitute \(r = 2\) back into the expression for the coefficient:

\(\binom{7}{2} \frac{9^2}{3^5}\)

\(= 21 \times \frac{81}{243}\)

\(= 21 \times \frac{1}{3}\)

\(= 7\)

Thus, the coefficient of \(x\) is 7.