= 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> is
$$
\boxed{(\tau_\gamma)_*(u)=u+([\gamma]\cdot u)[\gamma].}
$$
Reversing 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>.
\b[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 matrices
$$
A=\begin{pmatrix}1&1\\0&1\end{pmatrix},\qquad B=\begin{pmatrix}1&0\\-1&1\end{pmatrix},\qquad (AB)^3=-I,\quad(AB)^6=I.
$$
The <mapping class group> of the oriented closed <torus> is $SL_2(\mathbb Z)$, 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:
$$
A^{-1}=B(AB)^5,\qquad B^{-1}=(AB)^5A.
$$
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.
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