Strong operator topology is generated by the semi-norms $\rho_x(T):=\lVert Tx\rVert$.
What we have to show here is that if $T$ is linear and bounded, we can find $T_n$ of finite-rank such that $\rho_x(T-T_n)\to 0$ for all $x$.
Let $\{e_n\}$ an Hilbert basis of $H$, and $P_N$ the projection over $\operatorname{Span}\{e_1,\dots,e_N\}$. Let $T_N:=TP_N$. It's a finite-ranked operator. As $P_Nx\to x$ as $N\to +\infty$ and $T$ is bounded, we have convergence in the strong operator topology.
This result helps us to see that strong operator topology is really different from the topology induced by operator norm. Indeed, in the later, the closure of finite ranked operators is the set of compact operators, no matter whether the Hilbert space is separable or not.
The most usual way to show that a topology $A$ is stronger than a topology $B$ is to show that a convergent net in $A$ is also convergent in $B$. You show that the two topologies are equal by showing that both are stronger than the other.
Since all the topologies you want to consider are linear, it is enough to consider convergence at $0$.
Here, suppose that $T_j\to 0$ in norm, i.e. $\|T_j\|\to0$. For any $x\in H$,
$$
\|Tx\|+\|T^*x\|\leq\|T\|\,\|x\|+\|T^*\|\,\|x\|=2\|T\|\,\|x\|\to0;
$$
so norm convergence implies strong* convergence.
From $\|Tx\|\leq\|Tx\|+\|T^*x\|$ we get that strong* convergence implies strong convergence.
If $T_j\to0$ strongly, this means that $T_jx\to0$ for all $x\in H$. Then
$$
|\langle T_jx,y\rangle|\leq\|T_jx\|\,\|y\|\to0,
$$
so strong convergence implies weak convergence.
So far we have shown the "weaker than" implications. Now we need to see that they are strict.
Fix an orthonormal basis $\{e_j\}$ of $H$ and let $E_j$ be the projection onto the span of $e_j$ (i.e. $E_jx=\langle x,e_j\rangle e_j$). Then for any $x=\sum_j\alpha_je_j$,
$$
\|E_jx\|+\|E_j^*x\|=2\|E_jx\|=|\alpha_j|\to0,
$$
so $E_j\to0$ in the strong* topology. But $\|E_j\|=1$ for all $j$, so $\{E_j\}$ does not converge in norm.
Now consider the left-shift operator $T$ given by $Te_1=0$, $Te_j=e_{j-1}$ for $j>1$ (we don't need to assume $H$ separable here, just use a well-ordering of the index set). If you are familiar with it, $T$ is the adjoint of the unilateral shift. Put $T_n=T^n$, $n\in\mathbb N$. Then
$$
\|T_nx\|^2=\sum_{k>n}|\alpha_k|^2\to0\ \ \text{ as }n\to\infty;
$$
so $T_n\to0$ in the strong topology. But $T^*$ is an isometry, so
$$
\|T_nx\|+\|T_n^*x\|\geq\|T_n^*x\|=\|x\|
$$
for all $n$, so $\{T_n\}$ does not converge in the strong* topology.
Finally, we need a net that converges weakly but not strongly. Here we can use the unilaterial shift $S$ ($T^*$ above). If $x=\sum\alpha_ke_k$, $y=\sum_j\beta_je_j$, then
$$
|\langle S^nx,y\rangle|=|\sum_k\alpha_k\beta_{k+n}|\leq\sum_k|\alpha_k|\,|\beta_{k+n}|\leq\left(\sum_k|\alpha_k|^2\right)^{1/2}\,\left(\sum_k|\beta_{k+n}|^2\right)^{1/2}\to0
$$
(the second sum goes to zero because of the $n$). So $S^n\to0$ weakly but not strongly (recall that $S$ is an isometry).
Best Answer
This is an answer only to the first question.
Since $H$ is separable it has countable orthonormal basis $\{e_n:n\in\mathbb{N}\}$. Now for each $\lambda\in\ell_\infty$ consider diagonal operator $T_\lambda:H\to H$ well defined by equalities $T_\lambda(e_n)=\lambda_n e_n$. It is easy to check that $$ T_{\alpha' \lambda'+\alpha''\lambda''}=\alpha'T_{\lambda'}+\alpha''T_{\lambda''} $$ $$ \Vert T_\lambda\Vert=\Vert \lambda\Vert_\infty. $$ for all $\alpha',\alpha''\in\mathbb{C}$ and $\lambda',\lambda''\in\ell_\infty$. Thus we have an isometric inclusion $$ i:\ell_\infty\to\mathcal{B}(H):\lambda\to T_\lambda, $$ i.e. we can consider $\ell_\infty$ as subspace of $\mathcal{B}(H)$. Since $\ell_\infty$ is not separable then $\mathcal{B}(H)$ can't be separable too.