Everything from the 2 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.
Halogenoalkanes contain a polar Cδ+–Xδ− bond, so the carbon is attacked by nucleophiles. They are classed as primary, secondary or tertiary by how many carbons are attached to the C–X carbon. With aqueous NaOH they give alcohols; with KCN in ethanol, nitriles (one extra carbon); with ammonia in ethanol under pressure, amines: all nucleophilic substitution. With NaOH in ethanol they undergo elimination to alkenes. The rate of hydrolysis is governed by bond strength, not polarity: C–I is the weakest bond, so iodoalkanes react fastest.
Aqueous NaOH gives substitution (an alcohol); ethanolic NaOH gives elimination (an alkene). For the rate of hydrolysis, bond strength wins over bond polarity: iodoalkanes react fastest. KCN is the reagent that lengthens the carbon chain.
SN2 (substitution, nucleophilic, bimolecular) happens in one step: the nucleophile attacks the δ+ carbon from the side opposite the halogen while the C–X bond breaks, through a transition state with partial bonds. It is typical of primary halogenoalkanes, and its rate depends on both reactants. SN1 happens in two steps: the C–X bond breaks heterolytically (slowly) to give a carbocation, which the nucleophile then attacks (quickly). It is typical of tertiary halogenoalkanes, because the tertiary carbocation is stabilised by three alkyl groups and the crowded carbon blocks a backside attack. Its rate depends only on the halogenoalkane.
Curly arrows start from the lone pair on OH−, not from the minus sign, and the arrow from the C–X bond must point to the halogen. In SN2 draw the transition state in square brackets with partial bonds and an overall negative charge; SN2 has no carbocation.