Correct Answer: $\frac{61}{3}+\frac{17}{3 \sqrt{5}}$

Solution : Given: $\sin A =\frac{2}{3}$
We know that $\sin A=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{2}{3}$
We know, Hypotenuse ² = Base ² + Perpendicular ²
⇒ Base = $\sqrt{3^2-2^2}=\sqrt5$
We know, $\cos A =\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\sqrt{5}}{3}$
And,
⇒ $\tan A=\frac{\sin A}{\cos A}=\ \frac{2}{\sqrt{5}}$
Now,
$\left ( 7-\tan A \right )\left ( 3+\cos A \right )$
$=\left ( 7- \frac{2}{\sqrt{5}} \right )\left ( 3+\frac{\sqrt{5}}{3} \right )$
$=(\frac{7\sqrt5-2}{\sqrt{5}})(\ \frac{9+\sqrt{5}}{3})$
$=(\frac{63\sqrt5+35-18-2\sqrt5}{3\sqrt{5}})$
$=(\frac{61\sqrt5+17}{3\sqrt{5}})$
$=\frac{61}{3}+\frac{17}{3 \sqrt{5}}$
Hence, the correct answer is $\frac{61}{3}+\frac{17}{3 \sqrt{5}}$.