20. (1) Let the annual income at the end of each year be $A$ million yuan, with an annual profit of $10\%$, then:
the income from the first year's $A$ million yuan to the $n$-th year is $Q_{1}=A(1+0.1)^{n-1}$;
the income from the second year's $A$ million yuan to the $n$-th year is $Q_{2}=A(1+0.1)^{n-2}$;
the income from the $(n-1)$-th year's $A$ million yuan to the $n$-th year is $Q_{n-1}=A(1+0.1)$;
the income from the $n$-th year is $Q_{n}=A(1+0.1)^{0}=A$.
$Q_{1}, Q_{2} \cdots Q_{n}$ form a geometric sequence. Using the sum formula for a geometric sequence, the total income over these $n$ years is:
$$
Q=Q_{1}+Q_{2}+\cdots+Q_{n-1}+Q_{n}=\sum_{i=1}^{n} A(1+0.1)^{i-1}=\frac{A}{0.1}\left[(1+0.1)^{n}-1\right] \text {. }
$$

When $n=3$, $A=400$, $Q=\frac{400}{0.1} \cdot\left[(1+0.1)^{3}-1\right]=1324$ (million yuan). That is, the income after 3 years is 1324 million yuan, so 1300 million yuan can be obtained after 3 years.
(2) To recover the entire investment within 3 years, i.e., $Q=1600$, then $1600=\frac{A}{0.1}\left[(1+0.1)^{3}-1\right]$, solving for $A \approx 483.4$ (million yuan).
That is, an annual income of at least 483.4 million yuan is required to recover the entire investment within three years.
20. A research institute transfers a high-tech technology to a company for production. Both parties agree: the research institute will withdraw 4 million yuan from the company's profit at the end of each year and invest it in other investments, thus earning a profit of 10% annually.
（1）Can the research institute obtain a total income of 130 million yuan after three years?
（2）If the research institute requires a total income of 160 million yuan after three years, how much money should it withdraw from the company's profit at least each year?