Example. Solve the following problems:
1. Without using tables, find the value of $\lg ^{3} 2+\lg ^{3} 5+3 \lg 2 \cdot \lg 5$;
2. Simplify: $\frac{1-\operatorname{tg} \theta}{1+\operatorname{tg} \theta}$;
3. Solve the system of equations $\left\{\begin{array}{l}\frac{x}{y}+\frac{y}{x}=\frac{25}{12}, \\ x^{2}+y^{2}=7\end{array}\right.$
4. Given $\operatorname{tg} \theta=2$, find the value of $\frac{2}{3} \sin ^{2} \theta+\frac{1}{4} \cos ^{2} \theta$.
1. Original expression $=\mathbf{t g}^{3} 2+\lg ^{3} 5+3 \lg 2 \cdot 1 \mathrm{~g} 5$
$$
\begin{aligned}
& (\lg 2+\operatorname{Ig} 5) \\
= & (\lg 2+\lg 5)^{2}=1,
\end{aligned}
$$
2. Original expression $=\frac{\operatorname{tg} 45^{\circ}-\operatorname{tg} \theta}{1+\operatorname{tg} 45^{\circ} \operatorname{tg} \theta}$
$$
=\boldsymbol{\operatorname { t g }}\left(45^{\circ}-\theta\right) \text {; }
$$
3. Introduce auxiliary equation: $\frac{x}{y} \cdot \frac{y}{x},=1$,
and combine (1) and (3) to solve for $\frac{x}{y}=$ ?

Then combine (2) and (4) to solve.
$$
\text { 4. } \begin{aligned}
\because \operatorname{tg} \theta & =2, \therefore \frac{2}{3} \sin ^{2} \theta+\frac{1}{4} \cos ^{2} \theta \\
& =\frac{\frac{2}{3} \sin ^{2} \theta+\frac{1}{4} \cos ^{2} \theta}{\sin ^{2} \theta+\cos ^{2} \theta} \\
& =\frac{\frac{2}{3} \operatorname{tg}^{2} \theta+\frac{1}{4}}{\operatorname{tg}^{2} \theta+1}=\frac{\frac{2}{3} \times 2^{2}+\frac{1}{4}}{2^{2}+1}=\frac{7}{12}
\end{aligned}
$$