Let's write the following numerical expressions in a simpler form:

(1)

$$
\begin{gathered}
\frac{1}{(1+\sqrt{1+\sqrt{8}})^{4}}+\frac{1}{(1-\sqrt{1+\sqrt{8}})^{4}}+\frac{2}{(1+\sqrt{1+\sqrt{8}})^{3}}+\frac{2}{(1-\sqrt{1+\sqrt{8}})^{3}} \\
\frac{1}{(1+\sqrt{1+\sqrt{2}})^{4}}+\frac{1}{(1-\sqrt{1+\sqrt{2}})^{4}}+\frac{2}{(1+\sqrt{1+\sqrt{2}})^{3}}+\frac{2}{(1-\sqrt{1+\sqrt{2}})^{3}} \\
\frac{1}{(1+\sqrt{1+\sqrt{1.6}})^{4}}+\frac{1}{\left(1-\sqrt{1+\sqrt{1.6}}\right)^{4}}+\frac{2}{\left(1+\sqrt{1+\sqrt{1.6}}\right)^{3}}+ \\
\\
+\frac{2}{\left(1-\sqrt{1+\sqrt{1.6}}\right)^{3}}
\end{gathered}
$$
The three expressions are of the following form:

$$
\begin{aligned}
\frac{1}{(1+\sqrt{1+\sqrt{x}})^{4}}+ & \frac{1}{(1-\sqrt{1+\sqrt{x}})^{4}}+\frac{2}{(1+\sqrt{1+\sqrt{x}})^{3}}+ \\
& +\frac{2}{(1-\sqrt{1+\sqrt{x}})^{3}}
\end{aligned}
$$

By simplifying this, the desired forms can be obtained by substituting 8.2 and 1 for $x$. Moreover, this can be further simplified by introducing the notation

$$
\sqrt{1+\sqrt{x}}=y
$$

Let's bring the first two terms and the last two terms of (1) to a common denominator, choosing the product of the respective denominators as the common denominator, and immediately taking (2) into account. The common denominators are:

$$
\begin{aligned}
& (1+y)^{4}(1-y)^{4}=\left(1-y^{2}\right)^{4}=(-\sqrt{x})^{4}=x^{2} \\
& (1+y)^{3}(1-y)^{3}=\left(1-y^{2}\right)^{3}=(-\sqrt{x})^{3}=-x \sqrt{x}
\end{aligned}
$$

The sum of the two pairs of terms, after expanding the powers and taking (2) into account, is

$$
\begin{gathered}
\frac{1}{x^{2}}\left[(1-y)^{4}+(1+y)^{4}\right]=\frac{2}{x^{2}}\left(1+6 y^{2}+y^{4}\right)= \\
=\frac{2}{x^{2}}(1+6+6 \sqrt{x}+1+2 \sqrt{x}+x)=\frac{2}{x^{2}}(8+8 \sqrt{x}+x)
\end{gathered}
$$

and respectively

$$
-\frac{2}{x \sqrt{x}}\left[(1-y)^{3}+(1+y)^{3}\right]=-\frac{4}{x \sqrt{x}}\left(1+3 y^{2}\right)=-\frac{4}{x \sqrt{x}}(4+3 \sqrt{x})
$$

Now, taking $x^{2}$ as the common denominator, the entire expression transforms into

$$
\frac{2}{x^{2}}(8+8 \sqrt{x}+x-8 \sqrt{x}-6 x)=\frac{2}{x^{2}}(8-5 x)
$$

Accordingly, the values of the given expressions for $x=8$, $x=2$, and $x=1.6$ are -1, -1, and 0, respectively.