 Solution

Fix the algebraic intersection number of curves on an oriented surface by declaring $u\cdot v=+1$ when the ordered tangent pair $(u,v)$ agrees with the surface orientation. Orient an annular neighbourhood of $\gamma$ with coordinates $(\theta ,r)$ and orientation $d\theta \wedge dr$. A right-handed Dehn twist is represented there by $(\theta ,r)\mapsto (\theta +2\pi f(r),r)$, where $f$ increases from zero to one and is constant near the two boundary circles; it is the identity outside this annulus. A transverse arc gains one oriented copy of $\gamma$ per signed crossing. Consequently its homology action of a Dehn twist isReversing the orientation of $\gamma$ changes both factors' signs and leaves this expression unchanged. A separating curve has $[\gamma ]=0$, so its Dehn twist acts trivially on first homology.

The first assertion is true. Choose a basis $(a,b)$ for the first homology group of the torus with $a\cdot b=1$. The positive Dehn twists about these curves have matricesThe mapping class group of the oriented closed torus is $S{L}_2(\mathbb{Z})$, generated by $A,B$ and their inverses; the Euclidean algorithm on a primitive column gives this generation. The displayed relation rewrites those inverses as positive words:Replacing every inverse in a generating word proves the assertion. The closed torus hypothesis matters: a boundary twist is retained when a boundary circle must be fixed pointwise.