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}\\).