= Solution
By <irreducible traceless matrix Lie algebra is semisimple>, let $R$ be the <radical of a Lie algebra>[solvable radical]. <Lie's theorem> triangularizes its action, with finitely many diagonal <characters> $\lambda_i:R\to\mathbb C$, and supplies a nonzero common <eigenspace> for one of them. For $x\in L$, conjugation by $e^{tx}$ preserves $R$ and gives an equivalent <representation> of it. Thus the multiset of <characters> $\lambda_i\circ e^{t\operatorname{ad}x}$ is constant. Each <character> varies continuously in $t$ and has values in that finite set of <characters>, so it is individually constant. Differentiating at zero gives $\lambda_i([x,r])=0$ for all $r\in R$.
Choose a common <eigenspace> $V_\lambda\ne0$. For $v\in V_\lambda$,
$$
r(xv)=x(rv)+[r,x]v=\lambda(r)xv+\lambda([r,x])v=\lambda(r)xv.
$$
Thus $V_\lambda$ is $L$-invariant, so irreducibility makes it all of $\mathbb C^n$. Every element of $R$ acts as a scalar. Its <trace> is zero because $L\subseteq\mathfrak{sl}_n$, so that scalar is zero. Faithfulness of the <matrix> inclusion gives $R=0$. Hence $\boxed{L\text{ is semisimple}}$.
One sufficient diagonal-<matrix> condition for simplicity is that $L$ contain \b[all traceless <diagonal matrices>]. Indeed invariance under their adjoint actions then decomposes $L$ into that diagonal algebra and some <matrix>-unit spaces $\mathbb Ce_{ij}$. Put a directed edge $j\to i$ when $e_{ij}\in L$. Irreducibility makes the directed graph strongly connected; otherwise a terminal component spans a proper invariant coordinate subspace. Bracketing <matrix units> along a simple directed path supplies the <matrix unit> between its endpoints. Strong connectivity therefore supplies every off-<diagonal matrix> unit, so $L=\mathfrak{sl}_n$, which is simple. The mere existence of a <diagonal matrix> with distinct diagonal entries would not be a sufficient condition.
For the specified construction, $H$ has <dimension> two. Its diagonal adjoint action and the twelve distinct nonzero <weights of a representation>[weights] give a <direct sum> $L=H\oplus\bigoplus_{i=1}^{12}\mathbb Cg_i$, of <dimension> fourteen. The stated bracket condition and the <weight of a representation>[weight]-vector relations make it a <Lie algebra>. A subspace invariant under it is invariant under all the $g_i$ and hence their associative algebra, so it acts irreducibly on $\mathbb C^7$. The preceding result makes it semisimple.
Use <prime-dimensional faithful irreducible module forces simplicity>. To see this, decompose the algebra into simple <ideals of a Lie algebra>[ideals]. Complete reducibility for one <ideal of a Lie algebra>[ideal] and commutation with the others make its isotypic component invariant under the whole algebra, hence all of the <representation>. Writing that component as $W\otimes\operatorname{Hom}(W,V)$ and repeating produces a <tensor product> of irreducibles for the simple <ideals of a Lie algebra>[ideals]. Faithfulness makes every factor nontrivial, of <dimension> at least two. Their dimensions multiply to seven, so only one factor is possible. Thus the constructed $L$ is simple.
The diagonal algebra $H$ is toral, and its centralizer inside $L$ is just $H$, since no $g_i$ has zero <weight of a representation>[weight]. A maximal <toral subalgebra> containing $H$ lies in that centralizer, so equals $H$; it is a <Cartan subalgebra>. Thus the rank is two and the twelve <weights of a representation>[weights] are the <roots of a root system>[roots]. Pair them as $\alpha_{i+6}=-\alpha_i$. The opposite-<root of a root system>[root] brackets have <weight of a representation>[weight] zero and belong to $H$. For distinct $i,j\le6$, the sum cannot be zero, and the supplied <root of a root system>[root]-bracket result says their bracket is nonzero exactly when $\alpha_i+\alpha_j$ is a <root of a root system>[root]. This is equivalent to $-\alpha_i-\alpha_j$ being a <root of a root system>[root], proving the requested equivalence for the paired brackets.
Among rank-two <simple root> systems, $A_2,B_2,G_2$ have respectively six, eight and twelve <roots of a root system>[roots]. Therefore
$$
\boxed{L\text{ has type }G_2.}
$$
Choose <simple roots> $\alpha$ short and $\beta$ long. In $G_2$, $\beta+\alpha$ and $\beta+2\alpha$ are <roots of a root system>[roots]. Renumber with $g_1\in L_\alpha$, $g_3\in L_\beta$, $g_4\in L_{\beta+\alpha}$ and $g_5\in L_{\beta+2\alpha}$, retaining opposite partners in the other six positions. <Root spaces> are one dimensional, and the allowed bracket equality gives
$$
\boxed{[g_3,g_1]=\mu_1g_4,\qquad[g_4,g_1]=\mu_2g_5,\qquad\mu_1\mu_2\ne0.}
$$
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