Answer
(a) The magnitude of the net force is $4.94\times10^{-17}N$
(b) The magnitude of $F_2$ is $2.56\times10^{-17}N$
Work Step by Step
As $v_0=5.4\times10^5m/s, v=2.1\times10^6m/s$ and $x=0.038m$ (the distance the electron has traveled), the electron's acceleration can be calculated:
$$v^2=v_0^2+2ax$$ $$a=\frac{v^2-v_0^2}{2x}=5.42\times10^{13}m/s^2$$
(a) Since we now know $a=5.42\times10^{13}m/s^2$ and $m_e=9.11\times10^{-31}kg$, the net force acting on the electron can be found, using Newton's 2nd law:
$$F_{net}=m_e\times a=4.94\times10^{-17}N$$
(b) As $a\gt0$ and $\vec{F_{net}}=m_e\vec{a}$, $\vec{F_{net}}$ points in the $+x$ direction.
Since there are only 2 forces $F_1$ and $F_2$ acting on the electron and they point in opposite directions ($\vec{F_1}=+7.5\times10^{-17}N$ means the force points in the $+x$ direction) , we have $$F_1-F_2=F_{net}$$ $$F_2=F_1-F_{net}=(7.5\times10^{-17}N)-(4.94\times10^{-17}N)$$ $$F_2=2.56\times10^{-17}N$$
