H is a Hilber space

a) The way to show this is to show that if $A$ is normal that then $\ker(A)=\ker(A^*)$ holds and then to show that $\ker(A^*) = \mathrm{im}(A)^\perp$ always holds. For the first part note that
$$x\in  \ker(A) \iff \|Ax\|^2=0\iff\langle Ax,Ax\rangle =0$$
now for any $x\in H$:
$$\langle Ax,Ax\rangle  = \langle x , A^*Ax\rangle \overset*= \langle x, AA^*x\rangle = \langle A^*x,A^*x\rangle = \|A^*x\|^2$$
where $*$ uses that $A^*A=AA^*$, ie that $A$ is normal. So we find:
$$x\in \ker(A) \iff \langle Ax,Ax\rangle = 0 \iff \|A^*x\|^2=0\iff x\in \ker(A^*)$$
so $\ker(A)=\ker(A^*)$. Now we show $\ker(A^*)=\mathrm{im}(A)^\perp$. First we check $\ker(A^*)\subseteq \mathrm{im}(A)^\perp$, so let $x\in\ker(A^*)$ and $z=Ay\in\mathrm{im}(A)$, then:
$$\langle x, z\rangle = \langle x, Ay\rangle = \langle A^*x,y\rangle = 0$$
since $A^*x=0$. So for any $x\in \ker(A^*)$ you have for all $z\in \mathrm{im}(A)^\perp$ that $\langle x,z\rangle=0$ and as such $\ker(A^*)\subseteq \mathrm{im}(A)^\perp$. Next we check $\mathrm{im}(A)^\perp\subseteq \ker(A^*)$, here suppose $x\in \mathrm{im}(A)^\perp$ then:
$$\|A^*x\|^2=\langle A^*x,A^*x\rangle = \langle x,A(A^*x)\rangle = 0$$
since $A(A^*x)\in\mathrm{im}(A)$. Then necessarily $A^*x=0$ and $x\in\ker(A^*)$, giving $\mathrm{im}(A)^\perp\subseteq \ker(A^*)$.

So we have seen:
$$\ker(A)=\ker(A^*)=\mathrm{im}(A)^\perp$$

(b) This question is incomplete since its not clear what $h$ is. I think two situations are likely:
1. $h$ is a typo and should not appear in the question.
2. $h\in\Bbb K$ is some scalar.
Now if $h$ is a typo then you can "remove it" from the question by taking $h=1$ from point 2. As such I will simply assume that $h$ is some scalar in the base field and prove it for that case - if the question does not have any $h$ in it (ie the $h$ is a typo) then just take $h=1$ in what follows.

We will use the closed graph theorem, ie we will show that if $x_n\in H$ is a sequence so that $x_n\to x$ and $B x_n\to z$ for some $x,z\in H$ then $Bx=z$. This means that the graph of $B$ is closed - since $B$ has domain $H$ (ie a Banach space) the closed graph theorem gives that $B$ must be a bounded operator.

So let $y\in H$, one has:
$$\lim_n\langle B x_n,y\rangle = \langle \lim_n B x_n, y\rangle = \langle z,y\rangle$$
and also
$$\lim_n \langle B x_n, y\rangle = \lim_n \langle h x_n,By\rangle =\langle \lim_n hx_n,By\rangle = \langle hx,By\rangle = \langle Bx,y\rangle$$
so for all $y\in H$ one has $\langle Bx,y\rangle = \langle z,y\rangle$, ie
$$0=\langle Bx,y\rangle-\langle z,y\rangle = \langle Bx -z ,y\rangle$$
and then for $y=Bx-z$ you get
$$0=\langle Bx-z,Bx-z\rangle = \|Bx-z\|^2$$
and $Bx=z$. By the discussion above this verifies that $B$ has closed graph and since $B$ has domain all of $H$ we get that $B$ is bounded.