If you know that $\Phi(0)=1/2$, that answers part (d). You need
$$
\frac12=\Phi(0)=\Phi\left(\frac{\ln x-\mu}{\sigma}\right),
$$
so
$$
0 = \frac{\ln x-\mu}{\sigma}.
$$
Then remember that a fraction is $0$ only if the numerator is $0$.
Part (c) is really the main thing you should know about the log-normal distribution: $X$ is log-normally distributed iff $\ln X$ is normally distributed.
Two things suffice for part (c): (1) Differentiate both sides of the equality in (c). On one side you get $f(x)$. On the other side, recall that $\Phi'(x)=\frac{1}{\sqrt{2\pi}}\exp\left(\frac{-(x-\mu)^2}{2\sigma^2}\right)$, and also remember the chain rule.
If you know that $\Phi(0)=1/2$, that answers part (d). You need $$ \frac12=\Phi(0)=\Phi\left(\frac{\ln x-\mu}{\sigma}\right), $$ so $$ 0 = \frac{\ln x-\mu}{\sigma}. $$ Then remember that a fraction is $0$ only if the numerator is $0$.
Part (c) is really the main thing you should know about the log-normal distribution: $X$ is log-normally distributed iff $\ln X$ is normally distributed.
Two things suffice for part (c): (1) Differentiate both sides of the equality in (c). On one side you get $f(x)$. On the other side, recall that $\Phi'(x)=\frac{1}{\sqrt{2\pi}}\exp\left(\frac{-(x-\mu)^2}{2\sigma^2}\right)$, and also remember the chain rule.