Fix elements $\zeta$ and $\alpha$ with $\zeta$ a primitive third root of unity and $\alpha^3 = 3$. These generate a field $K = \Bbb Q(\zeta,\alpha)$ which is the splitting field of $x^3 - 3$, with Galois group $G$ the symmetric group on three letters.

Consider the elliptic curve $y^2 + 3y = x^3 - 3$. Unless I have miscalculated, the $3$-torsion points on this curve are the solutions to $18x^4 = 6x(2y+3)^2$. In particular, the points $(0,-1+\zeta)$ and $(\alpha,0)$ are independent 3-torsion points on this curve, so $K = \Bbb Q(E[3])$. Taking these as a basis, the resulting image of the Galois group into $GL_2(\Bbb F_3)$ must be $$\begin{bmatrix}* & * \\ 0 & 1\end{bmatrix}$$ because the elements $(\zeta^k \alpha, 0)$ are in the same Galois orbit and add to zero (they all lie on the line $y=0$).

Let $H < G$ be the subgroup of order three. Since the coefficient group $E[3]$ is $3$-torsion, a transfer argument implies that the restriction $H^1(G;E[3]) \to H^1(H;E[3])$ is injective with image the invariants under $G/H \cong \Bbb Z/2$.

If $$A = \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix},$$ then the group $H^1(H;E[3])$ is $ker(1 + A + A^2) / Im(1 - A)$, which is generated by the column vector $\left[\begin{smallmatrix}0 \\ 1\end{smallmatrix}\right]$. However, this element is fixed by the action of the order-2 subgroup of $G$ consisting of diagonal matrices, and hence represents a nonzero element of $H^1$.

Fix elements $\zeta$ and $\alpha$ with $\zeta$ a primitive third root of unity and $\alpha^3 = -4$. These generate a field $K = \Bbb Q(\zeta,\alpha)$ which is the splitting field of $x^3 + 4$, with Galois group $G$ the symmetric group on three letters.

Consider the elliptic curve $y^2 = x^3 + 1$. Unless I have miscalculated, the $3$-torsion points on this curve are the points $(x,y)$ with $x^4 + 4x = 0$. In particular, the points $(0,1)$ and $(\alpha,2\zeta+1)$ are independent 3-torsion points on this curve, so $K = \Bbb Q(E[3])$. Taking these as a basis, the resulting image of the Galois group into $GL_2(\Bbb F_3)$ must be $$\begin{bmatrix}1 & * \\ 0 & *\end{bmatrix}$$ because $(0,1)$ is fixed and the map must be injective.

Let $H < G$ be the subgroup of order three. Since the coefficient group $E[3]$ is $3$-torsion, a transfer argument implies that the restriction $H^1(G;E[3]) \to H^1(H;E[3])$ is injective with image the invariants under $G/H \cong \Bbb Z/2$.

If $$A = \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$$ represents the generator $\tau$ of $H$, then the group $H^1(H;E[3])$ is $ker(1 + A + A^2) / Im(1 - A)$, which is generated by the column vector $\left[\begin{smallmatrix}0 \\ 1\end{smallmatrix}\right]$. (This describes an element of $H^1$ by where an associated $1$-cocycle sends a chosen generator of $H$.)

As a $1$-cocycle, this is represented by the map $f:H \to E[3]$ with $$f(\tau^k) = (1 + A + \cdots + A^{k-1})\left[\begin{smallmatrix}0 \\ 1\end{smallmatrix}\right].$$ The action of the element $\sigma = \left[\begin{smallmatrix}1 & 0 \\ 0 & -1\end{smallmatrix}\right]$ on this cocycle is given by $$({}^\sigma f)(\tau) = \sigma \cdot f(\sigma^{-1} \tau \sigma) = \sigma \cdot f(\tau^2) = \left[\begin{smallmatrix}1 \\ 1\end{smallmatrix}\right]$$ which shows that the two $1$-cocycles ${}^\sigma f$ and $f$ represent the same element of $H^1$. Therefore, this element of $H^1(H;E[3])$ is invariant under $G/H$ and lifts to a nontrivial element of $H^1(G;E[3])$.