1. According to the Definition 3.48 (of a natural transformation), it is sufficient to specify each component of the natural transformation. As in Fredrick Eisele's figure (and the hint): compose the components, so \$$\forall c\in\mathcal{C}\,\,(\beta\circ\alpha)_c:=\beta_c\circ\alpha_c\$$. Since \$$\alpha_c\$$ and \$$\beta_c\$$ are morphisms, their composite exists and is a morphism (since \$$\mathcal{D}\$$ is a category), so \$$\beta\circ\alpha\$$ is well-defined. The only thing to check is naturality. \$$(\beta\circ\alpha)_d\circ F(f)\$$ \$$=\beta_d\circ\alpha_d\circ F(f)\$$ \$$=\beta_d\circ G(f)\circ\alpha_c\$$ \$$=H(f)\circ\beta_c\circ\alpha_c\$$ \$$=H(f)\circ(\beta\circ\alpha)_c\$$.

2. It is sufficient to define it on objects. To act like the identity, it must map an object to itself. \$$\forall F\in\mathcal{D}^\mathcal{C}\,\,(\mathrm{id}\_F)\_c(F(c))=F(c)\$$. This is a natural transformation since \$$(\mathrm{id}\_F)\_d\circ F(f)=\mathrm{id}\_{F(d)}\circ F(f)=F(f)=F(f)\circ\mathrm{id}\_{F(c)}=F(f)\circ(\mathrm{id}\_F)\_c\$$ (i.e., since each of its components is the identity morphism on the associated object, and unitality in \$$\mathcal{D}\$$). In the functor category \$$\mathcal{D}^\mathcal{C}\$$, unitality says \$$\forall\alpha\in\mathcal{D}^\mathcal{C}(F,G)\,\,\alpha\circ\mathrm{id\}_F=\alpha=\mathrm{id}\_G\circ\alpha\$$. The first equality follows from applying the above definition of equality on components in \$$\mathcal{C}\$$, and similarly for the second (the only difference being the functor for which the natural transformation is the identity), i.e., \$$\forall c\in\mathcal{C}\,\,(\alpha\circ\mathrm{id}\_G)\_c\$$ \$$=\alpha\_c\circ(\mathrm{id}\_G)\_c\$$ \$$=\alpha\_c\circ\mathrm{id}\_{G(c)}\$$ \$$=\alpha\_c\$$ \$$=\mathrm{id}\_{F(c)}\circ\alpha\_c\$$ \$$=(\mathrm{id}\_F)\_c\circ\alpha\_c\$$ \$$=(\mathrm{id}\_F\circ\alpha)\_c\$$.

Note that associativity in \$$\mathcal{D}^\mathcal{C}\$$ follows from associativity for morphisms in \$$\mathcal{D}\$$.