Hello Vanessa,
(a) We may model the amount $A$ in dollars in the account at time $t$ in years with the IVP:
$$\frac{dA}{dt}=rA$$ where $$A(0)=A_0$$
Separate variables, exchange dummy variables of integration and use boundaries as limits:
$$\int_{A_0}^{A(t)}\frac{du}{u}=r\int_0^t\,dv$$
Applying the FTOC, we find:
$$\left[\ln|u| \right]_{A_0}^{A(t)}=r\left[v \right]_0^t$$
$$\ln\left(\frac{A(t)}{A_0} \right)=rt$$
Convert from logarithmic to exponential form, solving for $A(t)$:
$$A(t)=A_0e^{rt}$$
Now, the average value in the account over a period of $t$ years is:
$$\overline{A(t)}=\frac{1}{t-0}\int_0^t A_0e^{ru}\,du=\frac{A_0}{rt}\int_0^t e^{ru}\,r\,du=\frac{A_0}{rt}\left[e^{ru} \right]_0^t=\frac{A_0\left(e^{rt}-1 \right)}{rt}$$
Now, using the given data:
$$A_0=77600,\,r=0.0535,\,t=10$$, we find:
$$\overline{A(10)}=\frac{77600\left(e^{0.0535\cdot10}-1 \right)}{0.0535\cdot10}\approx102613.05$$
(b) First, let's find the points of intersection:
$$f(x)=g(x)$$
$$-x^2+10x-6=4x+3$$
$$x^2-6x+9=0$$
$$(x-3)^2=0$$
You are correct, we find that $g(x)$ is tangent to $f(x)$ and so the area of region bounded by the two functions is zero.