For Equation-1
$$
\lim_{n\to \infty}P(|Y_n|< \epsilon)=\lim_{n\to \infty}P(-\epsilon <Y_n< \epsilon)=\lim_{n\to \infty}P(0<Y_n< \epsilon)
$$
Since the distribution of $Y_n$ is $1-(1-x)^n$, where $x$ is the distribution of $X_i$ (uniform distribution $(0,1)$), there is
$$
\lim_{n\to \infty}P(0<Y_n< \epsilon)=\lim_{n\to \infty}(1-(1-\epsilon)^n)=1
$$
For Equation-2
$$
\lim_{n\to \infty}P(|Z_n-1|< \epsilon)=\lim_{n\to \infty}P(1-\epsilon <Z_n<1+ \epsilon)=\lim_{n\to \infty}P(1-\epsilon<Z_n< 1)
$$
Since the distribution of $Z_n$ is $x^n$, where $x$ is the distribution of $X_i$ (uniform distribution $(0,1)$), there is
$$
\lim_{n\to \infty}P(1-\epsilon<Z_n< 1)=\lim_{n\to \infty}(1-(1-\epsilon)^n)=1
$$
For the cdf of the sample maximum we have
$$\begin{align*}
F_{X_{(n)}}(x)
&=\mathsf P(\text{max}{\{X_1,...,X_n}\}\leq x)\\\\
&=\mathsf P(X_1\leq x,...,X_n\leq x)\\\\
&=\mathsf P(X\leq x)^n\\\\
&=F_X(x)^n
\end{align*}$$
where
$$ F_{X}(x)=
\begin{cases}
0 & x \lt -1 \\
\frac{x+1}{2} & -1\leq x\leq1 \\
1 & x \gt 1
\end{cases} $$
Hence taking the derivative we get
$$f_{X_{(n)}} = \frac{n(x+1)^{n-1}}{2^n} I_{[-1,1]}(x)$$
Similarly for the sample minimum we have
$$\begin{align*}
F_{X_{(1)}}(x)
&=\mathsf P(\text{min}{\{X_1,...,X_n}\}\leq x)\\\\
&=1-\mathsf P(\text{min}{\{X_1,...,X_n}\}\gt x)\\\\
&=1-\left(1-F_X(x)\right)^n\\\\
&=1-\left(1-\frac{x+1}{2}\right)^n
\end{align*}$$
Hence taking the derivative we get
$$f_{X_{(1)}} = \frac{n\left(-x+1\right)^{n-1}}{2^n} I_{[-1,1]}(x)$$
Best Answer
The expectation operator is linear, hence it commutes with subtraction. For any random variables,
$$E(X-Y)=E(X)-E(Y),$$
provided these expectations exist.