For the second part, you may approximate $F=G=0$ near $(0,0,0)$ as follows,
$$F(x, y,z) \approx x+(1+y)-1 = x+y=0$$
$$G(x,y,z)\approx -z =0$$
Their normal vectors at origin are $(1,1,0)$ and $(0,0,-1)$ respectively. Thus, the tangent is $(1,1,0)\times (0,0,-1)=(-1,1,0)$ at the origin.
Remember the case of the real function of one real variable: $y=f(x)$. At a point $x_0$ where the function is differentiable, you have $f(x)=f(x_0)+f'(x_0)(x-x_0)+\text{ error term}$, where the error term is $o(x-x_0)$ when $x\to x_0$. This means that, close to $x_0$, the function is well-approximated by a linear function $f(x_0)+f'(x_0)(x-x_0)$.
For many variables, the situation is the same: if the function is differentiable at a point, it means that it can be closely approximated by a linear function. For example: let $F:X\to Y$ where $x\subseteq\mathbb R^n$ and $Y\subseteq\mathbb R^m$ ("$m$ functions of $n$ variables"), and let us write $F(x_1,\ldots,x_n)=(F_1(x_1,\ldots,x_n),\ldots,F_m(x_1,\ldots,x_n))$. If we assume that, at a point $(X_1,\ldots,X_n)\in X$ this function is differentiable, this means that:
$$F(x_1,\ldots,x_n)=F(X_1,\ldots,X_n)+dF_{(X_1,\ldots,X_n)}\left[x_1-X_1,\ldots,x_n-X_n\right]+\text{ error term}$$
where $dF$ (the differential of the function, taken at the point$ (X_1,\ldots,X_n)$) is actually a linear map, and the error term is "small" (for a suitable definition of "small") with respect to the vector $(x_1-X_1,\ldots,x_n-X_n)$. It just so happens that, in a one-function-of-one-variable, any linear map amounts to multiplying with a constant, which we call the derivative of the function, while here the derivative is more complicated but has the same nature.
Then, you learn later on the term "tangent surface": at the point $(X_1,\ldots,X_n)$, if you forget about the error term, you get a linear function that approximates well the original function near that point:
$$F(X_1,\ldots,X_n)+dF_{(X_1,\ldots,X_n)}\left[x_1-X_1,\ldots,x_n-X_n\right]$$
The image of this map (the "tangent surface") is an $n$-dimensional flat surface in $\mathbb R^m$ (hyperplane) which goes through $F(X_1,\ldots,X_n)$ just like the original function $F$, and is "close" to it in the neighbourhood of $(X_1,\ldots,X_n)$.
Also, you learn that, in the coordinates given, this map $dF$ has a matrix consisting of partial derivatives of the functions $F_1,\ldots,F_m$. Thus, any determinants of the square submatrices of that matrix are actually Jacobians.
The bigger point here is that you can use the machinery of the linear algebra to study the behaviour of the function $F$ near the chosen point. For example, when can you "invert" a linear map? You know that it depends on the rank of that linear map, and in particular, if the rank of it is $n$, then one of the $n\times n$ submatrices of the matrix of the linear map is nonzero. That immediately lets you invert the above linear map - for every choice of the other $m-n$ variables, you can solve for those $n$ variables coresponding to the (linearly independent) columns.
In effect: [1] you've replaced $F$ with its linear approximation, and [2] you know how to invert that approximation. The essence of the implicit function theorem is that then: you can then invert the original function $F$ - as long as you are close enough to the point $(X_1,\ldots,X_n)$ where you are running your analysis.
Best Answer
The main idea is that you have an equation that in theory could be solved for $y$ as a function of $x$ but the actual procedure is too hard/unknown. In other words, the equation gives enough information to uniquely specify a function but doesn't actually give it.
For example, $x^3y^3 + xy = 4$ is an equation cubic in $y$. With a lot of tedious work in algebra, it is possible to solve it and get $y$ alone. But if the equation is fifth degree or higher, then depending on the coefficients, there might not be any known way to find the formula directly.
It's a kind of specifying an answer by giving a problem that it solves. We call it indirect because it isn't "$y = $(blah that includes $x$'s but doesn't include any $y$'s)".