1.1 $\frac{\frac{1}{a}-\frac{1}{b+c}}{\frac{1}{a}+\frac{1}{b+c}}\left(1+\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right): \frac{a-b-c}{a b c}$;

$$
a=0.02, b=-11.05, c=1.07
$$

$1.2\left(\frac{1}{t^{2}+3 t+2}+\frac{2 t}{t^{2}+4 t+3}+\frac{1}{t^{2}+5 t+6}\right)^{2} \frac{(t-3)^{2}+12 t}{2}$.
1.1 Let's introduce the notations:

$$
\frac{\frac{1}{a}-\frac{1}{b+c}}{\frac{1}{a}+\frac{1}{b+c}}=A ; \quad 1+\frac{b^{2}+c^{2}-a^{2}}{2 b c}=B ; \quad \frac{a-b-c}{a b c}=C
$$

Then the given expression will take the form $A \cdot B: C$. In the expression $A$, the permissible values are $a \neq 0, b \neq -c$. Bringing the fractions to a common denominator, we get

$$
A=\frac{b+c-a}{b+c+a}
$$

For the expression $B$, using formula (1.8), we find

$$
B=\frac{(b+c)^{2}-a^{2}}{2 b c}=\frac{(b+c+a)(b+c-a)}{2 b c}
$$

Therefore,

$$
\begin{aligned}
& A \cdot B=\frac{b+c-a}{b+c+a} \cdot \frac{(b+c+a)(b+c-a)}{2 b c}=\frac{(b+c-a)^{2}}{2 b c} \\
& A \cdot B: C=\frac{(b+c-a)^{2}}{2 b c} \cdot \frac{a b c}{a-b-c}=\frac{a(a-b-c)}{2}
\end{aligned}
$$

Substituting the given values, we have

$$
\frac{0.02(0.02+11.05-1.07)}{2}=0.1
$$

Answer: 0.1.