2. Let the incircle $\odot O$ of $\triangle ABC$ intersect the median $BM$ on side $CA$ at points $G$ and $H$, with point $G$ between points $B$ and $H$. Given that $BG=HM$ and $AB=2$. What are the values of $BC$ and $CA$ when the length of segment $GH$ is maximized? Also, find the maximum value of $GH$.

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The translation is provided as requested, maintaining the original format and line breaks.
2. Suppose $BC > AB$. As shown in Figure 3, let $\odot O$ be tangent to sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$ respectively. Then $AE = AF$.
By the tangent-secant theorem, we have
$$
\begin{array}{l}
EM = \sqrt{HM(HM + GH)} \\
= \sqrt{BG(BG + GH)} = BF.
\end{array}
$$

Therefore, $AM = AB = 2$, $AC = 2AM = 4$.
Let $BC = 2a$. Since $BF = BD$, $CD = CE$, we have
$$
\begin{array}{l}
AE = AF = \frac{1}{2}(AB + AC - BC) = 3 - a, \\
EM = FB = a - 1.
\end{array}
$$

Let $GH = x$, $BG = HM = y$. Then
$$
y(x + y) = (a - 1)^2.
$$

By the Pythagorean theorem, the sum of the squares of the diagonals of a parallelogram equals the sum of the squares of its four sides. Therefore,
$$
\begin{array}{l}
BM^2 = \frac{1}{4}\left[2(AB^2 + BC^2) - AC^2\right] \\
= \frac{1}{4}\left[2(2^2 + 4a^2) - 4^2\right] = 2a^2 - 2, \\
[y + (x + y)]^2 = 2a^2 - 2.
\end{array}
$$

From equations (1) and (2), we get
$$
x = \sqrt{-2(a - 2)^2 + 2},
$$

where $4 - 2 < 2a < 4 + 2$, i.e., $1 < a < 3$.
Therefore, when $a = 2$, $x$ reaches its maximum value $\sqrt{2}$, i.e., when $AC = BC = 4$, the length of segment $GH$ reaches its maximum value $\sqrt{2}$.