= Solution
For the <Lie derivative of a tensor field>, the scalar rule follows because a <vector field> $X$ differentiates products of <smooth functions>. On <vector fields> the <Lie bracket of vector fields> satisfies
$$
[X,fY]=f[X,Y]+X(f)Y.
$$
This follows by applying both sides to an arbitrary <smooth function> and expanding the two compositions of derivations. Both the scalar and vector-field operators are real-linear, so the extension theorem applies and gives the unique contraction-compatible tensor operator $\mathcal L_X$.
For a type-$(1,1)$ tensor $A$, regard it as the pointwise <endomorphism> of $TM$ obtained by contracting its covector slot with a vector. The <endomorphism-induced tensor derivation> starts with
$$
D_Ah=0,\qquad D_AY=A(Y).
$$
It is real-linear and obeys $D_A(fY)=fA(Y)=fD_AY+(D_Af)Y$. The scalar operator zero is a derivation in every dimension, so there is no zero-dimensional obstruction here. Consequently it also extends uniquely. On a <differential one-form>, the two operators are
$$
\boxed{(\mathcal L_X\omega)(Y)=X(\omega(Y))-\omega([X,Y]),
\qquad D_A\omega=-\omega\circ A.}
$$
Their actions on arbitrary <tensor fields> follow by the tensor-product rule; $D_A$ adds $A$ in every contravariant slot and subtracts its dual action in every covariant slot.
Functions in this calculation belong to $C^\infty(M)$: the printed $C^\infty(X)$ in Q2(c) is a typographical error. The contraction $C^1_2$ pairs the sole covector with the new vector $Y$.
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