5. Let $a$ and $d$ be two positive integers. Construct a "triangle" similar to Pascal's triangle as follows:
$$
\begin{array}{l}
a \\
a+d \quad 2a \quad a+d \\
a+2d \quad 3a+d \quad 3a+d \quad a+2d \\
a+3d \quad 4a+3d \quad 6a+2d \quad 4a+3d \quad a+3d \\
\text{...... } \\
a+(n-1)d \\
\text{...... } \\
a+(n-1)d \\
\end{array}
$$
The construction of the "triangle" is as follows: the $n$-th row contains $n+1$ numbers, with the first and last numbers being $a+(n-1)d$, and the other numbers in the middle are obtained by adding the two adjacent numbers from the row above (for example: $4a+3d = (a+2d) + (3a+d)$, $6a+2d = (3a+d) + (3a+d)$, etc.).
(1) Prove that the sum of the numbers in the $n$-th row is $2^n a + (2^n - 2)d$;
(2) If the sum of the numbers in the $n$-th row is 1988, what is the maximum possible value of $n$? Find the corresponding values of $a$ and $d$, and justify your conclusion.
5. (1) Let the sum of the $n$-th row be $S_{n}$, then it is easy to find that $S_{n+1}=2 S_{n}+2 d$.

Given $S_{1}=2 a$,
then $S_{n+1}+2 d=2\left(S_{n}+2 d\right)=\cdots=2^{n}\left(S_{1}+2 d\right)=2^{n}(2 a+2 d)$.
Therefore, $S_{n}=2^{n} a+\left(2^{n}-2\right) d$. (Recursion)
(2) If $2^{n} a+\left(2^{n}-2\right) d=1988$,

then from $2^{n} a+\left(2^{n}-2\right) d=2^{n}(a+d)-2 d=1988$ we get $a+d \geqslant\left[\frac{1988}{2^{n}}\right]+1$,

thus $2^{n} a+\left(2^{n}-2\right) d=\left(2^{n}-2\right)(a+d)+2 a \geqslant\left(2^{n}-2\right)\left(\left[\frac{1988}{2^{n}}\right]+1\right)+2$.
When $n \geqslant 11$, $2^{n} a+\left(2^{n}-2\right) d \geqslant 2^{11}-2+2>1988$;
When $n=10$, $2^{n} a+\left(2^{n}-2\right) d \geqslant\left(2^{10}-2\right) \times 2>1988$;
When $n=9$, $2^{n} a+\left(2^{n}-2\right) d \geqslant\left(2^{9}-2\right) \times 4+2>1988$;
When $n=8$, $2^{n} a+\left(2^{n}-2\right) d \geqslant\left(2^{8}-2\right) \times 8+2>1988$;
When $n=7$, $2^{n} a+\left(2^{n}-2\right) d \geqslant\left(2^{7}-2\right) \times 16+2>1988$.
All lead to contradictions.
When $n=6$, $64(a+d)=1988+2 d$, we get $\quad a+d \geqslant 32$.
When $a+d=32$, we have $d=30 \Rightarrow a=2$.
When $a+d \geqslant 33$, $62(a+d)+2 a \geqslant 62 \times 33>1988$, contradiction!
In summary, the maximum value of $n$ is 6, at which $a=2, d=30$.