For your second question, you have $\mathbf{y}\sim N(\mathbf{X}\boldsymbol{\beta},\sigma^2 \mathbf{I})$ and suppose you're testing $\mathbf{C}\boldsymbol{\beta}=\mathbf{0}$. So, we have that (the following is all shown through matrix algebra and properties of the normal distribution -- I'm happy to walk through any of these details)
$ \mathbf{C}\hat{\boldsymbol{\beta}}\sim N(\mathbf{0}, \sigma^2 \mathbf{C(X'X)^{-1}C'}). $
And so,
$ \textrm{Cov}(\mathbf{C}\hat{\boldsymbol{\beta}})=\sigma^2 \mathbf{C(X'X)^{-1}C}. $
which leads to noting that
$ F_1 = \frac{(\mathbf{C}\hat{\boldsymbol{\beta}})'[\mathbf{C(X'X)^{-1}C'}]^{-1}\mathbf{C}\hat{\boldsymbol{\beta}}}{\sigma^2}\sim \chi^2 \left(q\right). $
You get the above result because $F_1$ is a quadratic form and by invoking a certain theorem. This theorem states that if $\mathbf{x}\sim N(\boldsymbol{\mu}, \boldsymbol{\Sigma})$, then $\mathbf{x'Ax}\sim \chi^2 (r,p)$, where $r=\textrm{rank}(A)$ and $p=\frac{1}{2}\boldsymbol{\mu}'\mathbf{A}\boldsymbol{\mu}$, iff $\mathbf{A}\boldsymbol{\Sigma}$ is idempotent. [The proof of this theorem is a bit long and tedious, but it's doable. Hint: use the moment generating function of $\mathbf{x'Ax}$].
So, since $\mathbf{C}\hat{\boldsymbol{\beta}}$ is normally distributed, and the numerator of $F_1$ is a quadratic form involving $\mathbf{C}\hat{\boldsymbol{\beta}}$, we can use the above theorem (after proving the idempotent part).
Then,
$ F_2 = \frac{\mathbf{y}'[\mathbf{I} - \mathbf{X(X'X)^{-1}X'}]\mathbf{y}}{\sigma^2}\sim \chi^2(n-p-1) $
Through some tedious details, you can show that $F_1$ and $F_2$ are independent. And from there you should be able to justify your second $F$ statistic.