5. Find the first decimal digit (immediately to the right of the decimal point) and the last digit before the decimal point (the units digit) of the number $(\sqrt{2}+\sqrt{3})^{1980}$, and prove your conclusion.
5. Let $a=(\sqrt{2}+\sqrt{3})^{1980}$ and $T_{n}=(\sqrt{2}+\sqrt{3})^{2 n}+(\sqrt{3}-\sqrt{2})^{2 n}=(5+2 \sqrt{6})^{n}+(5-2 \sqrt{6})^{n}$.
Since $5+2 \sqrt{6}, 5-2 \sqrt{6}$ are the roots of the equation $x^{2}=10 x-1$, according to Newton's formula, we have
$$
T_{n+2}=10 T_{n-1}-T_{n}(n=1,2, \cdots) \text { and } T_{1}=10, T_{2}=98 \text {. }
$$

From the recurrence relation, it is known that $T_{n}$ is an integer, and the unit digit of the sum of the unit digits of $T_{n+2}$ and $T_{n}$ is 0, so the unit digit of $T_{n+4}$ is the same as that of $T_{n}$, $T_{990}$ has the same unit digit as $T_{990-4 \times 247}=T_{2}=98$, i.e., the unit digit of $T_{990}$ is 8, because $0<5-2 \sqrt{6}<0.2$, so
$$
\begin{aligned}
0<(5-2 \sqrt{6})^{990} & <0.2^{990}<0.01^{330} \text {, hence } T_{990}=a+\underbrace{0.0 \cdots 0 p}_{\text {at least } 660 \text { zeros}} \\
T_{990} & =0<(5+2 \sqrt{6})^{990}+(5-2 \sqrt{6})^{990}=(\sqrt{2}+\sqrt{3})^{1980}+(\sqrt{3}-\sqrt{2})^{1980} .
\end{aligned}
$$

Therefore, the unit digit of $(\sqrt{2}+\sqrt{3})^{1980}$ is 7, and the first decimal place is 9 (and there are at least 660 consecutive 9s after the decimal point).