$f(x) = 0$ exactly when $\sin(5x) = \frac{k}{7}$ for an integer $-7 \le k \le 7;$ count solutions for each $k$ over five periods of $\sin(5x)$

Tangency to the $x$-axis also needs $f'(x) = 0,$ which forces $\cos(5x) = 0,$ i.e. $\sin(5x) = \pm 1$

$f(x) = 0$ exactly when $7\pi \sin(5x)$ is a multiple of $\pi,$ that is, $\sin(5x) = \frac{k}{7}$ for an integer $k$ with $|k| \le 7.$ As $x$ runs over $(0, 2\pi),$ the quantity $5x$ runs over $(0, 10\pi),$ five full periods. For $k = 0,$ the solutions are $5x = \pi, 2\pi, \ldots, 9\pi:$ $9$ values. For each of the $12$ values $k = \pm 1, \ldots, \pm 6,$ each period contributes $2$ solutions: $10$ values each. For $k = \pm 7,$ we need $\sin(5x) = \pm 1,$ which happens $5$ times each. So $n = 9 + 120 + 10 = 139.$

The graph is tangent to the $x$-axis at a zero exactly when $f'(x)$ $= 35\pi \cos(7\pi \sin(5x)) \cos(5x)$ $= 0$ there. At any zero, $\cos(7\pi \sin(5x)) = \cos(k\pi)$ $= \pm 1 \ne 0,$ so tangency requires $\cos(5x) = 0,$ which means $\sin(5x) = \pm 1:$ exactly the $10$ zeros with $k = \pm 7$ (there $\sin(5x)$ has an extremum, so $f$ touches without crossing). Thus $t = 10$ and $n + t = 149.$