The least count (L.C.) of a screw gauge is given by:

$$ \text{L.C.} = \frac{\text{Pitch}}{\text{Number of Divisions on Circular Scale}} $$

Let the original pitch be $$p$$ and the original number of divisions be $$n$$. Then, the original least count is:

$$ \frac{p}{n} = 0.01 \, \text{mm} $$

After modifications:

The pitch is increased by 75%, so the new pitch is:

$$ p_{\text{new}} = p + 0.75p = 1.75p $$

The number of divisions is reduced by 50%, so the new number of divisions is:

$$ n_{\text{new}} = 0.5n $$

The new least count is then:

$$ \text{L.C.}_{\text{new}} = \frac{p_{\text{new}}}{n_{\text{new}}} = \frac{1.75p}{0.5n} = \frac{1.75}{0.5} \times \frac{p}{n} = 3.5 \times \frac{p}{n} $$

Substitute the original least count:

$$ \text{L.C.}_{\text{new}} = 3.5 \times 0.01 \, \text{mm} = 0.035 \, \text{mm} $$

Expressed in the form $$\times 10^{-3} \, \text{mm}$$, this becomes:

$$ 0.035 \, \text{mm} = 35 \times 10^{-3} \, \text{mm} $$

Thus, the new least count is:

$$ 35 \times 10^{-3} \, \text{mm} $$