What does "maximum value" of a set of random variables mean?

I think it will help you to think through an example. Consider this one:

Let $x_1, x_2,$ and $x_3$ be three rolls with a fair die. Each $x_i$ can take the values from $1$ to $6$, however it is random which values they take.

Three Examples:

You roll $1$, $5$, and $2$ (i.e. $x_1 = 1$, $x_2 = 5$, $x_3 = 2$). Then $\max\{x_1,x_2,x_3\} = 5$.

You roll $1$, $1$, and $2$. Then $\max\{x_1,x_2,x_3\} = 2$.

You roll $3$, $2$, and $6$. Then $\max\{x_1,x_2,x_3\} = 6$.

Let us now calculate $\mathrm{Pr}(M_3<x)$ for, say, $x=4$. This means that all rolls must be below 4, i.e. 3 or lower. Each roll has $3/6 = 1/2$ chance of rolling $3$ or lower and the rolls are independent, hence the probability is given by $$\mathrm{Pr}(M_3<4) = \frac12\times\frac12\times\frac12 = \frac18.$$

In other words $$\mathrm{Pr}(M_3<4) = \left(\mathrm{Pr}(x_i<4)\right)^3. $$

If this example helps you understand it, it may help you generalize to other random variables.

The random variables $x_i$ are functions $\Omega\to\mathbb R$ where $\Omega$ is the sample space of the practicized probability space.

Then $M_n$ is also a function $\Omega\to\mathbb R$, and it is prescribed by:$$\omega\mapsto\max(x_1(\omega),\dots,x_n(\omega))$$

Note that: $$\{\omega\in\Omega\mid M_n(\omega)<x\}=\{\omega\in\Omega\mid x_1(\omega)<x\wedge\dots\wedge x_n(\omega)<x\}$$ so that also:$$\mathsf P(\{\omega\in\Omega\mid M_n(\omega)<x\})=\mathsf P(\{\omega\in\Omega\mid x_1(\omega)<x\wedge\dots\wedge x_n(\omega)<x\})$$or abbreviated:$$\mathsf P(M_n<x)=\mathsf P(x_1<x,\dots,x_n<x\}$$

The RHS equals: $$\mathsf P(x_1<x)\times\cdots\times\mathsf P(x_n<x)$$because the $x_i$ are independent.

If further the $x_i$ have equal distribution and $\text{Pr}(x):=\mathsf P(x_1<x)$ then we finally arrive at:$$=\text{Pr}(x)^n$$

Yes, it is the random variable with the highest value.

For your second question, consider $n$ real numbers $x_1, \dots, x_n$. Suppose the largest of these is less than some value $x$. Then it follows that $x_1, \dots, x_n$ must all be less than $x$ as well. By independence $$\mathbb{P}(M_n < x) = \mathbb{P}(X_1 < x, X_2 < x, \cdots , X_n < x) = \mathbb{P}(X_1 < x) \mathbb{P}(X_2 < x) \cdots \mathbb{P}(X_n < x)$$ and assuming that $X_1, \dots, X_n$ are all drawn from the same distribution (this was not stated in your question), it follows that $$\mathbb{P}(X_1 < x) \mathbb{P}(X_2 < x) \cdots \mathbb{P}(X_n < x) = [\mathbb{P}(X_1 < x)]^n$$ because all of $\mathbb{P}(X_1 < x)$, $\mathbb{P}(X_2 < x)$, ..., $\mathbb{P}(X_n < x)$ would be identical.