7.95 The diameter of a figure refers to the maximum distance between any two points in the figure. Given an equilateral triangle with a side length of 1, indicate how to cut it into two parts with a straight line, such that when these two parts are reassembled into a new figure, the new figure has the maximum diameter.
(1) If the new figure must be convex.
(2) Other cases.
[Solution] The maximum value of the diameter is $\frac{\sqrt{13}}{2}$.
If the cutting line passes through one vertex of the equilateral triangle and does not pass through the midpoint of the opposite side, there are three cases as shown in the figure.

Since the diameter is generated by the distance between two endpoints, only $B C^{\prime}$ and $D D^{\prime}$ are new lengths in these three figures. Since
$$
\begin{array}{l}
B C^{\prime} \leqslant \frac{1}{2}\left(B D+D C^{\prime}+B A+A C^{\prime}\right)=\frac{3}{2}180^{\circ}, \angle E C D+\angle A E D>180^{\circ}$, then $\triangle E D C$ cannot be attached to the side $A E$ of the quadrilateral $A E B D$, because in this case the resulting shape is not convex. Therefore, $\triangle E D C$ can only be attached to $B D$.
At this time, we have
$$
\begin{aligned}
\max \left\{A C^{\prime}, E C^{\prime}\right\} & \leqslant \frac{1}{2}\left(A B+B C^{\prime}+C^{\prime} D+D E+A E\right) \\
& =\frac{1}{2}(A B+B C+C A)=\frac{3}{2}<\frac{\sqrt{13}}{2} .
\end{aligned}
$$

In summary, the maximum value of the diameter of the convex figure formed by the pieces is $\frac{\sqrt{13}}{2}$.
(2) The maximum value of the diameter is 2.

Suppose a line divides $\triangle A B C$ into two parts $M, N$, and let $P$ be the common point of the two parts after they are joined together.
For any $X \in M, Y \in M$, we have
$X Y \leqslant X P+P Y \leqslant 1+1=2$.
Therefore, the diameter in this case does not exceed 2.
As shown in the figure, a figure with a diameter of 2 can be constructed.