= Solution
A <real (1, 1)-form> is a form $\varphi\in\Omega^{1,1}(X)$ satisfying $\bar\varphi=\varphi$. In holomorphic coordinates it has the form
$$
\varphi=i\sum_{j,k}h_{j\bar k}\,dz^j\wedge d\bar z^k,
\qquad h_{j\bar k}=\overline{h_{k\bar j}}.
$$
It is a <positive real (1, 1)-form> when
$$
-i\varphi(\xi,\bar\xi)>0
$$
for every nonzero tangent vector $\xi$ of type $(1,0)$, equivalently when the <Hermitian matrix> $(h_{j\bar k})$ is positive definite.
A <holomorphic local trivialization> of a <holomorphic line bundle> $L\to X$ is equivalently a nowhere-zero <holomorphic local frame> $e$. A connection $A$ is <unitary> when it preserves the fiberwise Hermitian inner product:
$$
d\,h(s,t)=h(\nabla^As,t)+h(s,\nabla^At).
$$
The <Chern connection> is the unique unitary connection whose $(0,1)$ part is the bundle's <Dolbeault partial connection> $\bar\partial_L$.
In a holomorphic frame, put $h=h(e,e)$. The <local formula for the Chern connection on a line bundle> is
$$
\nabla^Ae=(\partial\log h)e,
\qquad
F(A)=\bar\partial\partial\log h.
$$
The curvature therefore has type $(1,1)$. Since a unitary connection has imaginary curvature, $\overline{F(A)}=-F(A)$, and hence
$$
\overline{iF(A)}=iF(A).
$$
Thus $iF(A)$ is a real $(1,1)$-form.
For connections $A$ on $L$ and $\widehat A$ on $\widehat L$, the <tensor product connection> is defined on decomposable local sections by
$$
\nabla^{A\otimes\widehat A}(s\otimes\widehat s)
=\nabla^As\otimes\widehat s+s\otimes\nabla^{\widehat A}\widehat s.
$$
The <curvature of a tensor product connection> on line bundles is additive:
$$
F(A\otimes\widehat A)=F(A)+F(\widehat A).
$$
Consequently
$$
iF(A\otimes\widehat A)=iF(A)+iF(\widehat A),
$$
which is positive whenever both summands are positive.
For the final assertion, simultaneously diagonalize the positive Hermitian matrices of $\varphi$ and $\psi$ by congruence at the chosen point. In the resulting coframe,
$$
\varphi=i\sum_ja_jdz^j\wedge d\bar z^j,
\qquad
\psi=i\sum_jb_jdz^j\wedge d\bar z^j,
\qquad a_j,b_j>0.
$$
A direct wedge-product calculation gives
$$
(\varphi\wedge\psi)(\xi,\eta,\bar\xi,\bar\eta)
=\sum_{j<k}(a_jb_k+a_kb_j)
|\xi_j\eta_k-\xi_k\eta_j|^2.
$$
If $\xi$ and $\eta$ are linearly independent, at least one of these $2\times2$ minors is nonzero. Every coefficient is positive, so the sum is strictly positive. This is <wedge positivity for two positive (1, 1)-forms>.
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