This definition of concatenation is really just a more formal way of saying that in order to concatenate a string $w_1$ with a string $w_2$, we first append to $w_1$ the first character of $w_2$, then append to the result the second character of $w_2$, and continue in that fashion until we’ve exhausted $w_2$. For instance, it says that
$$\begin{align*}
abc\cdot def&=(abc\cdot de)f\\
&=\big((abc\cdot d)e\big)f\\
&=\big((abcd\cdot\lambda)e\big)f\\
&=\big((abcd)e\big)f\\
&=(abcde)f\\
&=abcdef\;.
\end{align*}$$
Note that it’s assumed that we know what it means to append a character to the end of a string; that’s what I did in the last two steps of the calculation.
I’ll prove by induction on the length of $w_2$ that the recursive definition really does tell you how to concatenate a string $w_1\in\Sigma^*$ with any string $w_2\in\Sigma^*$.
The basis step tells you how to concatenate a string $w_1$ with the string of length $0$, i.e., with the empty string: you just get $w_1$ back. Now suppose that you know how to concatenate $w_1$ with any string of length $n$; I claim that the induction step of the definition tells you how to concatenate $w_1$ with any string of length $n+1$. To see this, let $u$ be any string of length $n+1$. Then we can decompose $u$ into a string $w_2$ of length $n$ and a single character $x\in\Sigma$, so that $u=w_2x$. Then $$w_1\cdot w_2=w_1\cdot(w_2x)\;,$$ and the inductive step says that we can rewrite this as $$w_1\cdot w_2=w_1\cdot(w_2x)=(w_1\cdot w_2)x\;.$$ Since by hypothesis we already know how to concatenate $w_1$ with all strings of length $n$, we know how to form $w_1\cdot w_2$. It’s assumed from the beginning that we know what it means to append a single character to a string, so we also know what $(w_1\cdot w_2)x$ is, and therefore we know what $w_1\cdot u$ is. Finally, $u$ was an arbitrary string of length $n+1$, so we see that we now know how to concatenate $w_1$ with any string in $\Sigma^*$ of length $n+1$.
By induction, then, for each $n\in\Bbb N$ and each string $w_2\in\Sigma^*$ of length $n$ we know how to form $w_1\cdot w_2$. Finally, by definition every string in $\Sigma^*$ has length $n$ for some $n\in\Bbb N$, so we know how to form $w_1\cdot w_2$ for each $w_1\in\Sigma^*$.
This is one of those odd cases where we need to apply the definition of Kleene star directly. This should be definition 1.23 in your text. (I have the second edition, so the third edition could be different if you have that.) Below this definition there is a paragraph about Kleene Star. Since we can have any number of strings from the set concatenated together, we can have no strings (I.e the empty string). Every set Kleene starred contains the empty string.
Best Answer
A string belongs to the set $\{11\}^*\{01\}^*$ if and only if it can be written in the form $(11)^m(01)^n$ for some non-negative integers $m$ and $n$: it must be the result of concatenating $m$ copies of $11$ and $n$ copies of $01$ for some integers $m,n\ge 0$. The length of the string $(11)^m(01)^n$ is $2m+2n$, which is clearly always even. The length of the string $11101$, on the other hand, is odd, so $11101\notin\{11\}^*\{01\}^*$.
Showing that a string is not in a given set typically requires some sort of general argument, as in the previous paragraph, or a rather tedious syntactic analysis (see the addendum below). Showing that $11101\in\{111\}^*\{0\}^*\{01\}$, on the other hand, merely requires us to decompose $11101$ in a way that shows its membership in the set $\{111\}^*\{0\}^*\{01\}$. By definition every member of this set must have the form $(111)^m0^n01$ for some non-negative integers $m$ and $n$: it must be obtained by concatenating zero or more copies of $111$ with zero or more copies of $0$ and exactly one copy of $01$, in that order. Can we write $11101$ in that form? Sure: $11101=(111)^10^001$. That is, we can concatenate one copy of $111$, no copies of $0$, and the required single $01$ to form $11101$.
Added: It’s a bit tedious, but we can show that $11101\notin\{11\}^*\{01\}^*$ by thinking of $\{11\}^*\{01\}^*$ as a pattern, attempting to match $11101$ to that pattern, and showing that this can’t be done. It’s clear that if $\color{red}{11}101\in\{11\}^*\{01\}^*$, the $\color{red}{11}$ must belong to $\{11\}^*$. But the remaining $101$ starts with $10$, which belongs neither to $\{11\}^*$ nor to $\{01\}^*$, so there’s no way to continue the matching. Since the initial match of the $\color{red}{11}$ is forced, we’re stuck, and $11101$ cannot belong to $\{11\}^*\{01\}^*$.