Grade of the Hom functor 
Let $R$ be a Noetherian ring, $I$ an ideal and $M, N$ finite $R$-modules. Prove that $\operatorname{grade}(I,\operatorname{Hom}_R(M,N))\ge\min(2,\operatorname{grade}(I,N))$.

This question is Exercise 1.4.19 in the book of Winfried Bruns and Jürgen Herzog, Cohen-Macaulay Rings, Cambridge University Press, 1998.
I guess maybe Prop 1.2.9 in the book is helpful.
 A: As you observed, this is a consequence of the Depth Lemma. But first, another lemma:
$\DeclareMathOperator{grade}{grade}$
$\newcommand{Hom}{\operatorname{Hom}}$
$\newcommand{Ext}{\operatorname{Ext}}$
$\newcommand{coker}{\operatorname{coker}}$
Lemma: If $\grade(I,N) \ge 1$, then $\grade(I, \Hom_R(M,N)) \ge 1$ for any $R$-module $M$.
Proof: In general, $\grade(I, \_) \ge 1 \iff \Hom_R(R/I, \_) = 0$. By Hom-tensor adjointness, $\Hom_R(R/I, \Hom_R(M,N)) \cong \Hom_R(R/I \otimes_R M, N) \cong \Hom_R(M, \Hom(R/I, N))$.
Returning to your question: the result is clear if $\grade(I, N) = 0$, so we may assume $\grade(I,N) \ge 1$ (in fact the lemma also gives it for $\grade(I,N) = 1$, but this is merely fortunate).
Now, take a presentation of $M$,
$$0 \to K \to R^n \to M \to 0$$
Applying $\Hom_R(\_, N)$ and taking the long exact sequence yields
$$0 \to \Hom_R(M,N) \xrightarrow{\varphi} \Hom_R(R^n,N) \to \Hom_R(K,N) \to \Ext^1_R(M,N) \to 0$$
We can split this up into $2$ short exact sequences:
$$0 \to \Hom_R(M,N) \to N^n \to \coker \varphi \to 0$$
$$0 \to \coker \varphi \to \Hom_R(K,N) \to \Ext^1_R(M,N) \to 0$$
By the lemma above, $\grade(I, \Hom_R(K,N)) \ge 1$, so by the Depth Lemma for the second sequence, $\grade(I, \coker \varphi) \ge 1$, but then Depth Lemma for the first sequence gives the result (since $\grade(I,N) = \grade(I,N^n)$).
(Added for the sake of completeness: the statement of the Depth Lemma used is: if 
$$0 \to A \to B \to C \to 0$$ is an exact sequence of finite modules, then $\grade(I,A) \ge \min\{\grade(I,B), \grade(I,C) + 1\}$)
