F1 = 0.05 m, F2 = 0.1 m

Step 1: Understanding the situation
When an equiconvex lens is cut into two halves, it effectively turns into two plano-convex lenses. The original lens has a focal length (F) of 0.1 m. Each half will have different focal lengths due to their shape and geometry.

Step 2: Analyzing the original lens
The lens maker's formula is given by:
$$\frac{1}{F} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$
For an equiconvex lens, $R_1 = R$ and $R_2 = -R$ (since R2 is negative for the second surface), hence:
$$\frac{1}{F} = (n - 1) \left( \frac{1}{R} + \frac{1}{R} \right) = (n - 1) \left( \frac{2}{R} \right)$$
Which leads to:
$$F = \frac{R}{2(n - 1)}$$

Step 3: Analyzing the cut lens
Upon cutting, each half lens will behave as a plano-convex lens. For a plano-convex lens, the focal length can be given by:
$$F' = \frac{R}{2(n - 1)}$$ where R is the radius of curvature of the convex surface, which remains unchanged but changes the effective focal length.

Now, for each half:
For the plano-convex lens formed from one half: The radius of curvature will now effectively become half (R) because of how the thickness of the lens has halved.
Each new lens will have:
$$F' = \frac{1}{2} \left( \frac{R}{2(n - 1)} \right) = \frac{R}{4(n - 1)}$$

Step 4: Finding the equivalent focal lengths
Using the original data where F = 0.1 m, we know:
For the original lens $F = 0.1 m$, cutting the original lens into halves gives:
- The focal length of the plano-convex lens will be reduced by half.
- So the focal length of the first half lens will be 0.05 m.
- The other half lens will still maintain the original focal length of 0.1 m (since it hasn't changed geometrically in terms of its original curvature).

Final Result
Thus, the focal lengths of the new lenses are 0.05 m and 0.1 m.