Looks like you have the right idea. You divide the domain into sub-intervals, and evaluate in each sub-interval.
$f(x) + g(x) = \begin {cases} -3-6 = 7 & -3 \le x < -2.5\\-3-5 = 6 & -2.5 \le x < -2\\-2-4 = 5 & -2 \le x < -1.5\\&\vdots\end{cases}$
$f(x)g(x) = \begin {cases} (-3)(-6) = 18 & -3 \le x < -2.5\\(-3)(-5) = 15 & -2.5 \le x < -2\\(-2)(-4) = 8 & -2 \le x < -1.5\\&\vdots\end{cases}$
If $x \in [0,1)$, $f(x) = \int_0^x \lfloor t \rfloor \, dt= \int_0^x 0\, dt=0$.
If $x \in [1,2)$, $f(x) = \int_0^1 \lfloor t\rfloor \, dt + \int_1^x \lfloor t \rfloor \, dt= \int_0^1 0\, dt + \int_1^x 1\, dt=x-1$.
If $x \in [2, 3)$, $f(x) = \int_0^1 \lfloor t\rfloor \, dt + \int_1^2 \lfloor t\rfloor \, dt + \int_2^x \lfloor t \rfloor \, dt= \int_0^1 0\, dt + \int_1^2 1\, dt + \int_2^x 2\, dt=1+2(x-2)$.
In general, if $x \in [n, n+1), n \in \mathbb{N} \cup \{0\}$,
\begin{align}
f(x) &= \sum_{i=0}^{n-1} \int_i^{i+1} \lfloor t \rfloor \, dt + \int_n^x \lfloor t \rfloor \, dt \\
&= \sum_{i=0}^{n-1} \int_i^{i+1} i \, dt + \int_n^x n \, dt \\
&= \left(\sum_{i=0}^{n-1} i \right)+ n(x-n)
\end{align}
Try to simplify the last term.
Best Answer
Use that you define $[x]$ such that $[x]\leq x<[x]+1$. Now replace this into all the statements you want to prove.