1. The calculation goes as follows: \begin{align} d_A \circ d_A(\alpha) & = d_A(d\alpha + [A, \alpha]) \\ & = d(d\alpha + [A,\alpha]) + [A, d\alpha + [A,\alpha]] \\ & = d^2 \alpha + d[A, \alpha] + [A, d\alpha] + [A, [A, \alpha]] \\ & = 0 + [dA, \alpha] - [A, d\alpha] + [A, d\alpha] + \tfrac{1}{2}[[A, A], \alpha] \\ & = [dA + \tfrac{1}{2}[A,A], \alpha] \\ & = [d_A A, \alpha]. \end{align} There's a couple basic identities you need to check in the process, but it's nothing difficult. The $\tfrac{1}{2}$ shows up when you use the Jacobi identity, so it seems that perhaps Audin is missing it, but I don't have access to the book currently and can't look at what she does. For matrix Lie groups, $\tfrac{1}{2}[A, A] = A \wedge A$ where we consider $A \wedge A$ as a matrix product, so sometimes you will see $F(A) = dA + A \wedge A$ in the context of matrix Lie groups.

2. Simply use the definition of the derivative in an affine space (since $\mathcal{A}$ is affine): \begin{align} (T_A F)(\phi) & = \lim_{t \to 0} \frac{1}{t} (F(A + t\phi) - F(A)) \\ & = \lim_{t \to 0} \frac{1}{t}(d_{A + t\phi} (A + t\phi) - d_A A) \\ & = \lim_{t \to 0} \frac{1}{t}(d(A + t\phi) + \tfrac{1}{2}[A, A + t\phi] - dA - \tfrac{1}{2}[A,A]) \\ & = \lim_{t \to 0} \frac{1}{t}(dA + td\phi + \tfrac{1}{2}[A,A] + \tfrac{1}{2}t[A, \phi] - dA - \tfrac{1}{2}[A,A]) \\ & = d\phi + \tfrac{1}{2}[A, \phi] \\ & = d_A \phi. \end{align}

1. The calculation goes as follows:

\begin{align} d_A \circ d_A(\alpha) & = d_A(d\alpha + [A, \alpha]) \\ & = d(d\alpha + [A,\alpha]) + [A, d\alpha + [A,\alpha]] \\ & = d^2 \alpha + d[A, \alpha] + [A, d\alpha] + [A, [A, \alpha]] \\ & = 0 + [dA, \alpha] - [A, d\alpha] + [A, d\alpha] + \tfrac{1}{2}[[A, A], \alpha] \\ & = [dA + \tfrac{1}{2}[A,A], \alpha] \\ & = [d_A A, \alpha]. \end{align}

There's a couple basic identities you need to check in the process, but it's nothing difficult. The $\tfrac{1}{2}$ shows up when you use the Jacobi identity, so it seems that perhaps Audin is missing it, but I don't have access to the book currently and can't look at what she does. For matrix Lie groups, $\tfrac{1}{2}[A, A] = A \wedge A$ where we consider $A \wedge A$ as a matrix product, so sometimes you will see $F(A) = dA + A \wedge A$ in the context of matrix Lie groups.

2. Simply use the definition of the derivative in an affine space (since $\mathcal{A}$ is affine):

\begin{align} (T_A F)(\phi) & = \lim_{t \to 0} \frac{1}{t} (F(A + t\phi) - F(A))\\ & = \lim_{t \to 0} \frac{1}{t}(d_{A + t\phi} (A + t\phi) - d_A A)\\ & = \lim_{t \to 0} \frac{1}{t}(d(A + t\phi) + \tfrac{1}{2}[A, A + t\phi] - dA - \tfrac{1}{2}[A,A])\\ & = \lim_{t \to 0} \frac{1}{t}(dA + td\phi + \tfrac{1}{2}[A,A] + \tfrac{1}{2}t[A, \phi] - dA - \tfrac{1}{2}[A,A])\\ & = d\phi + \tfrac{1}{2}[A, \phi]\\ & = d_A\phi. \end{align}

1. The calculation goes as follows: \begin{align} d_A \circ d_A(\alpha) & = d_A(d\alpha + [A, \alpha]) \\ & = d(d\alpha + [A,\alpha]) + [A, d\alpha + [A,\alpha]] \\ & = d^2 \alpha + d[A, \alpha] + [A, d\alpha] + [A, [A, \alpha]] \\ & = 0 + [dA, \alpha] - [A, d\alpha] + [A, d\alpha] + [[A, A], \alpha] \\ & = [dA + [A,A], \alpha] \\ & = [d_A A, \alpha]. \end{align} There's a couple basic identities you need to check in the process, but it's nothing difficult.

2. Simply use the definition of the derivative in an affine space (since $\mathcal{A}$ is affine): \begin{align} (T_A F)(\phi) & = \lim_{t \to 0} \frac{1}{t} (F(A + t\phi) - F(A)) \\ & = \lim_{t \to 0} \frac{1}{t}(d_{A + t\phi} (A + t\phi) - d_A A) \\ & = \lim_{t \to 0} \frac{1}{t}(d(A + t\phi) + [A, A + t\phi] - dA - [A,A]) \\ & = \lim_{t \to 0} \frac{1}{t}(dA + td\phi + [A,A] + t[A, \phi] - dA - [A,A]) \\ & = d\phi + [A, \phi] \\ & = d_A \phi. \end{align}

1. The calculation goes as follows: \begin{align} d_A \circ d_A(\alpha) & = d_A(d\alpha + [A, \alpha]) \\ & = d(d\alpha + [A,\alpha]) + [A, d\alpha + [A,\alpha]] \\ & = d^2 \alpha + d[A, \alpha] + [A, d\alpha] + [A, [A, \alpha]] \\ & = 0 + [dA, \alpha] - [A, d\alpha] + [A, d\alpha] + \tfrac{1}{2}[[A, A], \alpha] \\ & = [dA + \tfrac{1}{2}[A,A], \alpha] \\ & = [d_A A, \alpha]. \end{align} There's a couple basic identities you need to check in the process, but it's nothing difficult. The $\tfrac{1}{2}$ shows up when you use the Jacobi identity, so it seems that perhaps Audin is missing it, but I don't have access to the book currently and can't look at what she does. For matrix Lie groups, $\tfrac{1}{2}[A, A] = A \wedge A$ where we consider $A \wedge A$ as a matrix product, so sometimes you will see $F(A) = dA + A \wedge A$ in the context of matrix Lie groups.

2. Simply use the definition of the derivative in an affine space (since $\mathcal{A}$ is affine): \begin{align} (T_A F)(\phi) & = \lim_{t \to 0} \frac{1}{t} (F(A + t\phi) - F(A)) \\ & = \lim_{t \to 0} \frac{1}{t}(d_{A + t\phi} (A + t\phi) - d_A A) \\ & = \lim_{t \to 0} \frac{1}{t}(d(A + t\phi) + \tfrac{1}{2}[A, A + t\phi] - dA - \tfrac{1}{2}[A,A]) \\ & = \lim_{t \to 0} \frac{1}{t}(dA + td\phi + \tfrac{1}{2}[A,A] + \tfrac{1}{2}t[A, \phi] - dA - \tfrac{1}{2}[A,A]) \\ & = d\phi + \tfrac{1}{2}[A, \phi] \\ & = d_A \phi. \end{align}