45. Find the largest real number $\alpha$, such that there exists an infinite sequence of positive integers $\left\{a_{n}\right\}_{n=1}^{+\infty}$, satisfying:
(1) For any $n \in \mathbf{N}^{*}$, we have $a_{n}>2008^{n}$;
(2) For any $n \in \mathbf{N}^{*}, n \geqslant 2$, the number $a_{n}^{\alpha}$ does not exceed the greatest common divisor of all numbers in the set $\left\{a_{i}+a_{j} \mid i+j=n, i, j \in \mathbf{N}^{*}\right\}$.
45. First, prove $\alpha \leqslant \frac{1}{2}$.

In fact, let $\left\{a_{n}\right\}$ be a sequence of positive integers satisfying conditions (1) and (2). We have: for any $0 < \varepsilon < 1$, there are infinitely many positive integers $n$ such that
$$a_{n}^{2-\varepsilon} \leqslant a_{2 n}.$$
If (11) does not hold, then there exists a positive integer $n$ such that
$$a_{n}^{2-\varepsilon} > a_{2 n}.$$
Since $a_{n} \geqslant 2008^n$, we have
$$a_{2 n} \geqslant 2008^{2 n}.$$
Thus,
$$a_{n}^{2-\varepsilon} > 2008^{2 n},$$
which implies
$$a_{n} > 2008^{\frac{2 n}{2-\varepsilon}}.$$
Since $a_{n} \geqslant 2008^n$, we have
$$2008^n > 2008^{\frac{2 n}{2-\varepsilon}},$$
which simplifies to
$$n > \frac{2 n}{2-\varepsilon},$$
or
$$2-\varepsilon > 2,$$
which is a contradiction. Therefore, (11) holds.

Now, let $n$ be a positive integer satisfying (11). Then,
$$a_{n}^{2-\varepsilon} \leqslant a_{2 n},$$
so
$$a_{n}^{(2-\varepsilon) \alpha} \leqslant a_{2 n}^{\alpha} \leqslant \operatorname{gcd}\left\{a_{i}+a_{j} \mid i+j=2 n, i, j \in \mathbf{N}^{*}\right\} \leqslant 2 a_{n},$$
which implies
$$a_{n}^{(2-\varepsilon) \alpha - 1} \leqslant 2.$$
Thus,
$$2008^{n((2-\varepsilon) \alpha - 1)} \leqslant 2.$$
Since there are infinitely many such $n$, we have
$$(2-\varepsilon) \alpha - 1 \leqslant 0,$$
or
$$\alpha \leqslant \frac{1}{2-\varepsilon}.$$
By the arbitrariness of $\varepsilon$, we conclude that
$$\alpha \leqslant \frac{1}{2}.$$

Next, prove that $\alpha = \frac{1}{2}$ is achievable by constructing a sequence $\left\{a_{n}\right\}$ that satisfies the conditions. Use the Fibonacci sequence $\left\{F_{n}\right\}$, which satisfies: $F_{1} = F_{2} = 1$, $F_{n+2} = F_{n+1} + F_{n}$, $n = 0, 1, 2, \cdots$. Then,
$$F_{n} = \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{n} - \left(\frac{1-\sqrt{5}}{2}\right)^{n}\right),$$
and by its recurrence relation, we have: when $m > n$,
$$F_{m} = F_{m-n} F_{n+1} + F_{m-n-1} F_{n}.$$

Notice that $\frac{1+\sqrt{5}}{2} > 1$ and $-1 < \frac{1-\sqrt{5}}{2} < 0$. Therefore, for sufficiently large $n$,
$$F_{n} > 2008^n.$$
Now define the sequence $\left\{a_{n}\right\}$ as $a_{n} = 3 F_{2m}$, $n = 1, 2, \cdots$. Condition (1) is clearly satisfied. For condition (2), we have
$$\begin{array}{l}
F_{2i} = F_{t(i+j)} F_{t(i-j)+1} + F_{t(i+j)-1} F_{t(i-j)}, \\
F_{2tj} = F_{t(i+j)} F_{t(j-i)-1} + F_{t(i+j)+1} F_{t(j-i)},
\end{array}$$
where we extend $\left\{F_{n}\right\}_{n=1}^{+\infty}$ to negative indices using the recurrence relation $F_{n+2} = F_{n+1} + F_{n}$. This gives us $F_{-2m} = -F_{2m}$ and $F_{-(2m-1)} = F_{2m-1}$, $m = 1, 2, \cdots$. Since $t$ is even, we have
$$\begin{aligned}
F_{2t} + F_{2tj} & = 2 F_{t(i+j)} F_{t(i-j)+1} + (F_{t(i+j)+1} - F_{t(i+j)-1}) F_{t(j-i)} \\
& = F_{t(i+j)} (2 F_{t(i-j)+1} + F_{t(j-i)}).
\end{aligned}$$
Thus, $F_{t(i+j)} \mid (F_{2i} + F_{2tj})$, and therefore $F_{nt} \mid (F_{2i} + F_{2tj})$ when $i + j = n$. This implies
$$\operatorname{gcd}\left\{a_{i} + a_{j} \mid i + j = n\right\} \geqslant 3 F_{m}.$$
Now,
$$a_{n} = 3 F_{20} = 3 F_{on} (F_{or+1} + F_{m-1}) \leqslant 9 F_{m}^2 \leqslant \left(\operatorname{gcd}\left\{a_{i} + a_{j} \mid i + j = n\right\}\right)^2,$$
so
$$a_{n}^{\frac{1}{2}} \leqslant \operatorname{gcd}\left\{a_{i} + a_{j} \mid i + j = n\right\}.$$
In conclusion, the maximum real number $\alpha = \frac{1}{2}$.