Uniform topology is determined by single operator norm $\Vert\cdot\Vert$ given by $\Vert T\Vert=\sup\{\Vert Tx\Vert: x\in\operatorname{B}_X\}$.
Let $(T_i:i\in I)$ be any net convergent to $T$ in the uniform topology, then we have
$$
\lim_{i}\Vert T_i^*-T^*\Vert
=\lim_{i}\Vert (T_i-T)^*\Vert
=\lim_{i}\Vert T_i-T\Vert=0
$$
Hence $^*:\mathcal{B}(H)\to\mathcal{B}(H)$ is continuous in the uniform topology.
Weak topology is determined by family of seminorms $\{\Vert\cdot\Vert_{x,y}:x,y\in H\}$ given by $\Vert T\Vert_{x,y}=|\langle Tx,y\rangle|$. Let $(T_i:i\in I)$ be any net convergent to $T$ in the weak topology. For any $x,y\in H$ we have
$$
\lim_{i}\Vert T_i^*-T^*\Vert_{x,y}
=\lim_{i}|\langle(T_i^*-T^*)x,y\rangle|
=\lim_{i}|\langle x,(T_i-T)y\rangle|\\
=\lim_{i}\Vert T_i-T\Vert_{y,x}
=0
$$
Hence $^*:\mathcal{B}(H)\to\mathcal{B}(H)$ is continuous in the weeak topology.
Strong topology is determined by family of seminorms $\{\Vert\cdot\Vert_x:x\in H\}$ given by $\Vert T\Vert_x=\Vert Tx\Vert$. For a given $x,y\in H$ we define the operator $x\bigcirc y:H\to H:z\mapsto \langle z,y\rangle x$. One can easily check that $(x\bigcirc y)^*=y\bigcirc x$. Let $\{e_n:n\in\mathbb{N}\}\subset H$ be an orthonormal family. For any $x\in H$ we get
$$
\lim_{n}\Vert(e_1\bigcirc e_n)\Vert_{x}
=\lim_{n}|\langle x, e_n\rangle|
=0
$$
Note that the last equality is the consequence of Bessel's inequality.
On the other hand
$$
\lim_{n}\Vert(e_1\bigcirc e_n)^*\Vert_{x}
=\lim_{n}\Vert(e_n\bigcirc e_1)\Vert_{x}
=|\langle x, e_1\rangle|
$$
which is not $0$ in general. Thus the map $^*:\mathcal{B}(H)\to\mathcal{B}(H)$ is not even sequentially strongly continuous.
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
The topologies of function spaces are often described in terms of convergence. For example, the compact-open can be described in terms of uniform convergence whenever the underlying space has a uniform structure.
You should know about how weak convergence is so-called because it is much weaker than strong convergence: for example, in any separable Hilbert space, the orthonormal basis vectors weakly converge to $0$ but do not strongly tend to $0$. In $L^2$, the functions $\chi_{[n,n+1]}$ tend to zero weakly but not strongly.
$T_\alpha\to T$ in the SOT iff. for all $x\in X$ and all $\varepsilon\gt0$, there exists $A$ such that $\|(T_\alpha-T)x\|\lt\varepsilon$ for all $\alpha\gt A$. This can be seen by inspecting the basis. The convergence occurs in the WOT iff. for all $x,\varepsilon$ and for all continuous linear functionals $\psi$, $|\psi((T_\alpha-T)x)|\lt\varepsilon$. These are, fixing $x$, precisely the definitions of $T_\alpha x\to Tx$ strongly and $T_\alpha x\rightharpoonup Tx$ weakly; since strong implies weak but not the other way round, these are (in infinite spaces) strictly stronger and weaker conditions. Then if we assert that this should be true for all $x$ we get $T_\alpha\to T$ strongly and $T_\alpha\leftharpoonup T$ weakly as strictly stronger and weaker conditions, and the different topologies topologise the respective notions of convergence. However, the SOT is still “weak” in that the norm topology on $L(X,Y)$ is stronger: that topologises uniform convergence, since it requires that the maximum value of the difference $(T_\alpha-T)$ be bounded for all $x$, with the same bound.
I hope this interprets your question correctly. If you want a clear example, consider $X,Y=\ell^2$ and let $T_n((x_k)_{k\in\Bbb N})=x_n$, which is a continuous linear operator. $T_n\rightharpoonup0$ weakly, so, in the WOT, and also in the SOT, but not in the norm topology.