Nyquist Sampling Rate Problem
The solution is correct, here's how I prefer to solve it.

Understand that if we can correctly identify frequency A, and A is greater than frequency B, then you can also correctly identify frequency B.

Here, \$25\pi t\$ is greater than \$10\pi t\$ so we can ignore the \$10\pi t\$ and only focus on the \$25\pi t\$.

Delays doesn't affect the frequency, so we can ignore the \$+9\$ in the \$y(t)\$ function.

At \$t=1\$, one second has passed and we can read the data straight out from the \$e^{i25 \pi t}\$, if we plug in the \$3\$ from \$y(t)=x(3t)\$ we get \$e^{i75 \pi}\$

One revolution is \$ 2\pi\$, this means we'll divide \$75\pi\$ by \$2\pi\$ to get \$37.5 \text{ Hz}\$. And then we multiply \$37.5 \text{ Hz}\$ by \$2\$ for the Nyquist frequency which is 75 Hz => 75 samples / sec.