$A\left(t\right)=80000\cdot {1.06}^t$

Do it.

Estimate: about $190000$, the calculator gives you $A\left(15\right)\approx 191725$.

See table.

t	0	1	2	3	4	5	6
log(N)	1.70	1.92	2.15	2.37	2.60	2.83	3.05

Yes, you get a straight line that goes through $(0;1.70)$ and $(4;2.60))$. Since the graph of $log\left(N\left(t\right)\right)$ is approximated by a straight line, then $N\left(t\right)$ must be an exponential function (roughly).

$log\left(N\right)\approx 1.70+0.22t$

$N\left(t\right)\approx {10}^{1.70+0.22t}={10}^{1.70}\cdot {\left({10}^{0.22}\right)}^t\approx 50\cdot {1.66}^t$

For $V\left(t\right)=b\cdot g^t$ the following is true: $V\left(0\right)=2=b\cdot g^0$ and $V\left(5\right)=6=b\cdot g^5$.
This results in: $b=2$ and $g^5=\frac{6}{2}=3$, and therefore $g\approx 1.25$. A suitable formula is $V\left(t\right)\approx 2\cdot {1.25}^t$.

$V\left(t\right)=10$ gives you ${1.25}^t=5$ and therefore $t={}^{1.25}log\left(5\right)\approx 7.21$.

$V\left(t\right)=1$ gives you ${1.25}^t=0.5$ and therefore $t={}^{1.25}log\left(0.5\right)\approx 3.11$.

The tick labels on both axes are powers of $10$.

$log\left(m\right)\approx 1.1$ and $log\left(P\right)\approx 2.4$.

$log\left(P\right)=a\cdot log\left(m\right)+b$ through $(1.1;2.4)$ and $(2.9;2.0)$.
This results in $a=\frac{-0.4}{1.8}\approx -0.22$ and $b\approx 2.64$, so $log\left(P\right)\approx -0.22\cdot log\left(m\right)+2.64$.

$P\approx {10}^{-0.22\cdot log\left(m\right)+2.64}={\left({10}^{log\left(m\right)}\right)}^{-0.22}\cdot {10}^{2.64}\approx 440\cdot m^{-0.22}$.