From tensor calculus, we know that
\begin{equation} g^{\mu\nu}=\delta_{\lambda}^{\mu}\delta_{\phi}^{\nu}g^{\lambda\phi}.\tag{1} \end{equation} Based on (1), is the following true? \begin{equation} g^{\mu\nu}g_{\lambda\phi}^{}=\delta_{\lambda}^{\mu}\delta_{\phi}^{\nu}.\tag{2} \end{equation}
The simplest possible concrete example demonstrates this to be wrong. Take $[\eta]=\text{diag}(-1,+1,+1,+1)$, then:
$$\eta^{00}\eta_{11}=-1\cdot +1=-1\tag{1},$$
on the other hand:
$$\delta^0_1\delta^0_1=0. \tag{2}$$
No. The first relation is a sum over $\lambda$ and $\phi$. You can't multiply the inverse metric $g^{\lambda\phi}$ on the RHS with anything to get the LHS.
No it wouldn't. First of all you can't have more than two indices repeating which happens when you multiplied with $g_{\lambda \phi}$, this is your biggest mistake. To confirm what you can get by this multiplication, multiply it by $g_{\alpha \beta}$
$$g^{\mu \nu} g_{\alpha \beta} = \delta^{\mu}_{\lambda} \delta^{\nu}_{\phi} g^{\lambda \phi} g_{\alpha \beta}$$
$$g^{\mu \nu} g_{\alpha \beta} = \delta^{\mu}_{\lambda} g^{\lambda \nu} g_{\alpha \beta}$$
$$g^{\mu \nu} g_{\alpha \beta} = g^{\mu \nu} g_{\alpha \beta}$$
